import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns
import sklearn.datasets as data
%matplotlib inline
sns.set_context('poster')
sns.set_style('white')
sns.set_color_codes()
plot_kwds = {'alpha' : 0.5, 's' : 80, 'linewidths':0}Clustering Notes
Clustering Notes
The following notes follow along this blog post. Clustering methods all use a central idea of a distance or metric defined by a function \(d(i,j)\). It is often the Euclidean distance.
DBSCAN
Concepts
Consider a set of points in some space to be clustered. Let \(X = \{x_1, ..., x_n\}\), a set of \(n\) objects, \(\epsilon\) be a parameter specifying a radius of a neighborhood with respect to some point. Points in DBSCAN fall into 3 categories: - a point \(p\) is a core point if there are at least a minimum number of points within \(\epsilon\) of it, including itself - a point \(q\) is directly reachable from point \(p\) if \(q\) is within distance \(\epsilon\), but only if \(p\) is a core point - a point \(q\) is reachable form \(p\) if there is a path \(p_1, ..., p_n\) with \(p_1 = p\) and \(p_n = q\) where each \(p_{i+1}\) is directly reachable from \(p_i\). This implies that the initial point and all points on the path must be core points, with the potential exception of \(q\) - all points not reachable from any other point are outliers or noise points
Reachability is not a symmetric relation, so another concept of connection is required for the extent of clusters: - two points are density-connected if there is a point \(o\) such that both \(p\) and \(q\) are reachable from \(o\)
Clusters satisfy two properties: - all points within a cluster are mutually density-connected - if a point is density-reachable from some point of the cluster, it is part of the cluster as well
Steps
Choose minimum points \(m_{pts}\) threshold and \(\epsilon\)
Find the points in the \(\epsilon\) neighborhood of every point, and identify the core points with more than \(m_{pts}\) neighbors
Find the connected components of core points on the neighbor graph, ignoring all non-core points
Assign each non-core point to a nearby cluster if the cluster is an \(\epsilon\) neighbor, otherwise assign it to noise
Optimization Criterion
For any possible clustering \(C = \{C_1, ..., C_t\}\) out of the set of all clusterings \(\mathcal{C}\), DBSCAN minimizes the number of clusters such that every pair of points, \(p, q\) in a cluster is density-reachable.
\[\underset{C \subset\mathcal{C}, d_{db}(p, q) \le \epsilon \forall p, q \in C_i \forall C_i \in C}\min |C|\]
HDBSCAN
This is a hierarchical extension of DBSCAN. For a chosen \(m_{pts}\)
Concepts
Core Distance \(d_{core}(x_p)\)
applies to object \(x_p \in X\), with respect to \(m_{pts}\), is defined as the distance from \(x_p\) to its \(m_{pts}\)-nearest neighbor, including itself
\(\epsilon\)-Core Object
Any object \(x_p \in X\) is where the following holds: \[d_{core}(x_p) \le \epsilon\]
Mutual Reachability Distance \(d_{mreach}(x_p, x_q)\)
is defined as \[d_{mreach}(x_p, x_q) \equiv\max\{d_{core}(x_p), d_{core}(x_q), d(x_p, x_q) \}\]
Mutual Reachability Graph \(G_{m_{pts}}\)
The complete graph in which the objects of \(X\) are vertices and the weight of each edge is the mutual reachability distance (wrt \(m_{pts}\)) between the respective pair of objects.
Subgraphs of \(G_{m_{pts}}\) and Hierarchies
If we take a subgraph \(G_{m_{pts}, \epsilon}\) obtained by removing all edges from \(G_{m_{pts}}\) with weights greater than \(\epsilon\), then the connected components are the \(\epsilon\)-core objects in the DBSCAN clusters, and remaining objects noise. Thus, DBSCAN partitions for \(\epsilon \in [0, \infty)\) can be produced in a hierarchical way: at 0 we have the finest clusters, or leaves of a dendrogram, while at higher values, we approach the root of the dendrogram.
Steps
Core Algorithm
- Compute core distance wrt \(m_{pts}\) for all data objects in \(X\)
- Compute minimum spanning tree (MST) of \(G_{m_{pts}}\), the mutual reachability graph
- Extend the MST to obtain \(MST_{ext}\) by adding for each vertex, a “self edge” with the core distance of the corresponding object as weight
- Extract the HDBSCAN hierarchy as a dendrogram from \(MST_{ext}\)
- 4.1 For the root of the tree, assign all objects the same label
- 4.2 Iteratively remove all edges from the \(MST_{ext}\) in decreasing order of weight, with ties being removed simultaenously
- 4.2.1 Before each removal, set the dendrogram scale value of the current hierarchical level as the weight of the edge(s) to be removed
- 4.2.2 After each removal, assign labels to connected component(s) that contain the end vertices of removed edges, a new cluster label to a component if it still has at least one edge, else noise
Customizable Cluster Processing Final Step
When reaching a lower \(\epsilon\) level in the tree, we must consider what happens to the clusters. We can define a separate minimum cluster size which can be applied to the following: - if all cluster’s subcomponents are spurious, it disappears - if only one of cluster’s subcomponents is not spurious, keep original label, ie cluster shrinkage - if two or more of cluster’s subcomponents are not spurious, “true” cluster split
Optimal Clustering
Cluster Stability
When increasing the density threshold, we can see prominent clusters remain and shrink or split while others disappear. A cluster can be seen as a set of points that whose density \(f(x)\), exceed a threshold, ie. \(\lambda = 1/{\epsilon}\). For a density contour cluster \(C_i\) that appears at density level \(\lambda_{min}(C_i)\), we can formalize its stability by defining its excess of mass as:
\[E(C_i) = \int_{x \in C_{i}}\Big( f(x) - \lambda_{min}(C_i)\Big)dx\]
This exhibits monotonic behavior along branches of the dendrogram, so it cannot be used to compare stabilities of nested clusters. Instead, we can define the relative excess of mass:
\[E_R(C_i) = \int_{x \in C_{i}}\Big( \lambda_{max}(x, C_i) - \lambda_{min}(C_i)\Big)dx\]
where \(\lambda_{max}(C_i)\) is the density level at which \(C_i\) is split or disappears, and \(\lambda_{max}(x, C_i) = \min\{f(x), \lambda_{max}(C_i)\}\)
For an HDBSCAN hierarchy, where we have finite data set \(X\), cluster labels, and desnity thresholds associated with each hierarchical level, we can adapt the previous expression to define the stability of a cluster \(C_i\) as:
\[ S(C_i) = \sum_{x_j \in C_i} \Big( \lambda_{max}(x, C_i) - \lambda_{min}(C_i)\Big) = \sum_{x_j \in C_i} \Big(\frac{1}{\epsilon_{min}(x_j, C_i)} - \frac{1}{\epsilon_{max}(C_i)}\Big)\]
where \(\lambda_{min}(C_i)\) is the minimum density level at which \(C_i\) exists, \(\lambda_{max}(x_j, C_i)\) is the density level beyond which object \(x_j\) no longer belongs to cluster \(C_i\) and \(\epsilon_{max}(C_i)\) and \(\epsilon_{min}(x_j, C_i)\) are the corresponding values for the threshold \(\epsilon\).
Cluster Selection Algorithm
With the notion of cluster stability developed, we can have extraction of the most prominent clusters formulated as an optimization problem of maximizing cluster stabilities subject to constraints:
\[\underset{\delta_2, ..., \delta_{\kappa}}\max J = \sum^\kappa_{i=2}\delta_iS(C_i)\]
\[\text{subject to} \begin{cases} \delta_i \in \{0,1\}, i=2,...,\kappa \\ \sum_{j \in I_h} \delta_j = 1, \forall h \in L\\ \end{cases}\]
where \(\delta_i\) indicates if cluster \(i\) is in the flat solution, \(L = \{h | C_h \text{is a leaf cluster}\}\) is the set of indices of leaf clusters, and \(I_h = \{j | j \ne 1 \text{ and }C_j \text{ is an ascendant of } C_h\}\), the set of indices of all clusters on the path from \(C_h\) (included) to the root (excluded). The constraints prevent nested clusters on the same path from being selected at the same time.
Selecting the clusters requires bottom up processing of nodes excluding the root, starting with the leaves, deciding whether to keep node \(C_i\) or a best so far selection of clusters in its subtrees. In the process, the total stability, \(\hat{S}(C_i)\) is updated:
\[\hat{S}(C_i) = \begin{cases} S(C_i), &\text{if } C_i \text{ is a leaf node}\\ \max\{S(C_i), \hat{S}(C_{i_l}) + \hat{S}(C_{i_r}) &\text{if } C_i \text{ is an internal node}\\ \end{cases} \]
Steps
- Initialize \(\delta_2 = ... \delta_\kappa = 1\) and for all leaf nodes, set \(\hat{S}(C_h) = S(C_h)\)
- Starting from the deepest levels, do bottom-up (excluding root):
- 2.1 If \(S(C_i) < \hat{S}(C_{i_r}) + \hat{S}(C_{i_l})\), set \(\hat{S}(C_i) = \hat{S}(C_{i_l}) + \hat{S}(C_{i_r})\) and set \(\delta_i=0\)
- 2.2 Else, set \(\hat{S}(C_i) = S(C_i)\) and set \(\delta_{(.)} = 0\) for all clusters in \(C_i\)’s subtrees.
Implementation
Hierarchical clustering can typically be slow but thanks to parallelization, can be sped up via GPUs. Nvidia’s blog entry, Faster HDBSCAN Soft Clustering with RAPIDS cuML, shows comparisons with multiple methods on datasets of varying sizes.
moons, _ = data.make_moons(n_samples=50, noise=0.05)
blobs, _ = data.make_blobs(n_samples=50, centers=[(-0.75,2.25), (1.0, 2.0)], cluster_std=0.25)
test_data = np.vstack([moons, blobs])
plt.scatter(test_data.T[0], test_data.T[1], color='b', **plot_kwds)
import hdbscan
clusterer = hdbscan.HDBSCAN(min_cluster_size=5, gen_min_span_tree=True)
clusterer.fit(test_data)HDBSCAN(gen_min_span_tree=True)In a Jupyter environment, please rerun this cell to show the HTML representation or trust the notebook.
On GitHub, the HTML representation is unable to render, please try loading this page with nbviewer.org.
HDBSCAN(gen_min_span_tree=True)
clusterer.minimum_spanning_tree_.plot(edge_cmap='viridis',
edge_alpha=0.6,
node_size=80,
edge_linewidth=2)
clusterer.single_linkage_tree_.plot(cmap='viridis', colorbar=True)
clusterer.condensed_tree_.plot()
clusterer.condensed_tree_.plot(select_clusters=True, selection_palette=sns.color_palette())
palette = sns.color_palette()
cluster_colors = [sns.desaturate(palette[col], sat)
if col >= 0 else (0.5, 0.5, 0.5) for col, sat in
zip(clusterer.labels_, clusterer.probabilities_)]
plt.scatter(test_data.T[0], test_data.T[1], c=cluster_colors, **plot_kwds)
Soft Clustering for HDBSCAN
Concept
- Hard clustering assigns a single cluster label or noise
- Soft clustering assigns a vector of probabilities
Instructional Implementation
import hdbscan
import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns
import pandas as pd
import matplotlib as mpl
from scipy.spatial.distance import cdist
%matplotlib inline
sns.set_context('poster')
sns.set_style('white')
sns.set_color_codes()
plot_kwds={'alpha':0.25, 's':60, 'linewidths':0}
palette = sns.color_palette('deep', 12)data = np.load('clusterable_data.npy')
fig = plt.figure()
ax = fig.add_subplot(111)
plt.scatter(data.T[0], data.T[1], **plot_kwds)
ax.set_xticks([])
ax.set_yticks([]);
len(data)2309
# build a clustering object to fit the data
clusterer = hdbscan.HDBSCAN(min_cluster_size=15).fit(data)
clusterer.labels_array([ 5, 5, 5, ..., -1, -1, 5], dtype=int64)
len(clusterer.labels_)2309
clusterer.probabilities_array([1. , 0.85883269, 0.90828071, ..., 0. , 0. ,
1. ])
clusterer.probabilities_array([1. , 0.85883269, 0.90828071, ..., 0. , 0. ,
1. ])
# visualize clustering using scores as saturation
pal = sns.color_palette('deep', 8)
colors = [sns.desaturate(pal[col], sat) for col, sat in zip(clusterer.labels_,
clusterer.probabilities_)]
plt.scatter(data.T[0], data.T[1], c=colors, **plot_kwds)
Distance Based Membership
This takes advantage of features to examine clustering further, as pointed out in these docs. In particular, we use the CondensedTree object, obtained via `clusterer. which is a dendrogram, or family tree of various clusters.
_raw_treeis the tree in the form of a numpy recarray allowing lookup based on field names where each row represents an edge between a parent and child clusters:- parent: id of parent cluster
- child: id of child cluster
- lambda_val: inverse distance, aka density at which edge forms
- child_size: number of points in the child cluster
def exemplars(cluster_id, condensed_tree):
raw_tree = condensed_tree._raw_tree
# Just the cluster elements of the tree, excluding singleton points
clusterer_tree = raw_tree[raw_tree['child_size'] > 1]
# Get the leaf cluster nodes under the cluster we are considering
leaves = hdbscan.plots._recurse_leaf_dfs(clusterer_tree, cluster_id)
# Now collect up the last remaining points of each leaf cluster (the heart of the leaf)
result = np.array([])
for leaf in leaves:
max_lambda = raw_tree['lambda_val'][raw_tree['parent'] == leaf].max()
points = raw_tree['child'][(raw_tree['parent'] == leaf) &
(raw_tree['lambda_val'] == max_lambda)]
result = np.hstack((result, points))
return result.astype('int32')
tree = clusterer.condensed_tree_
plt.scatter(data.T[0], data.T[1], c='grey', **plot_kwds)
for i, c in enumerate(tree._select_clusters()):
c_exemplars = exemplars(c, tree)
plt.scatter(data.T[0][c_exemplars], data.T[1][c_exemplars], c=palette[i], **plot_kwds)C:\Users\Jonathan\AppData\Local\Temp\ipykernel_16588\4108267181.py:20: UserWarning: *c* argument looks like a single numeric RGB or RGBA sequence, which should be avoided as value-mapping will have precedence in case its length matches with *x* & *y*. Please use the *color* keyword-argument or provide a 2D array with a single row if you intend to specify the same RGB or RGBA value for all points.
plt.scatter(data.T[0][c_exemplars], data.T[1][c_exemplars], c=palette[i], **plot_kwds)

def min_dist_to_exemplar(point, cluster_exemplars, data):
dists = cdist([data[point]], data[cluster_exemplars.astype(np.int32)])
return dists.min()
def dist_vector(point, exemplar_dict, data):
result = {}
for cluster in exemplar_dict:
result[cluster] = min_dist_to_exemplar(point, exemplar_dict[cluster], data)
return np.array(list(result.values()))
def dist_membership_vector(point, exemplar_dict, data, softmax=False):
if softmax:
result = np.exp(1./dist_vector(point, exemplar_dict, data))
result[~np.isfinite(result)] = np.finfo(np.double).max
else:
result = 1./dist_vector(point, exemplar_dict, data)
result[~np.isfinite(result)] = np.finfo(np.double).max
result /= result.sum()
return result
exemplar_dict = {c:exemplars(c, tree) for c in tree._select_clusters()}
colors = np.empty((data.shape[0], 3))
for x in range(data.shape[0]):
membership_vector = dist_membership_vector(x, exemplar_dict, data)
color = np.argmax(membership_vector)
saturation = membership_vector[color]
colors[x] = sns.desaturate(pal[color], saturation)
plt.scatter(data.T[0], data.T[1], c=colors, **plot_kwds);C:\Users\Jonathan\AppData\Local\Temp\ipykernel_16588\1075745609.py:16: RuntimeWarning: divide by zero encountered in divide
result = 1./dist_vector(point, exemplar_dict, data)

Outlier Based Membership
- these are density-based memberships using cluster persistence as a baseline of comparison
- GLOSH algorithm compares cluster persistence with how long a point stayed in the cluster
- we can compare cluster persistence with the merge height of the point with a fixed cluster in the dendrogram
- merge height represents the disimilarity between two clusters
- we can do that with multiple clusters to form a vector which can be normalized into memership scores
def max_lambda_val(cluster, tree):
cluster_tree = tree[tree['child_size'] > 1]
leaves = hdbscan.plots._recurse_leaf_dfs(cluster_tree, cluster)
max_lambda = 0.0
for leaf in leaves:
max_lambda = max(max_lambda,
tree['lambda_val'][tree['parent'] == leaf].max())
return max_lambda
def points_in_cluster(cluster, tree):
leaves = hdbscan.plots._recurse_leaf_dfs(tree, cluster)
return leaves
def merge_height(point, cluster, tree, point_dict):
cluster_row = tree[tree['child'] == cluster]
cluster_height = cluster_row['lambda_val'][0]
if point in point_dict[cluster]:
merge_row = tree[tree['child'] == float(point)][0]
return merge_row['lambda_val']
else:
while point not in point_dict[cluster]:
parent_row = tree[tree['child'] == cluster]
cluster = parent_row['parent'].astype(np.float64)[0]
for row in tree[tree['parent'] == cluster]:
child_cluster = float(row['child'])
if child_cluster == point:
return row['lambda_val']
if child_cluster in point_dict and point in point_dict[child_cluster]:
return row['lambda_val']
def per_cluster_scores(point, cluster_ids, tree, max_lambda_dict, point_dict):
result = {}
point_row = tree[tree['child'] == point]
point_cluster = float(point_row[0]['parent'])
max_lambda = max_lambda_dict[point_cluster] + 1e-8 # avoid zero lambda vals in odd cases
for c in cluster_ids:
height = merge_height(point, c, tree, point_dict)
result[c] = (max_lambda / (max_lambda - height))
return result
def outlier_membership_vector(point, cluster_ids, tree,
max_lambda_dict, point_dict, softmax=True):
if softmax:
result = np.exp(np.array(list(per_cluster_scores(point,
cluster_ids,
tree,
max_lambda_dict,
point_dict
).values())))
result[~np.isfinite(result)] = np.finfo(np.double).max
else:
result = np.array(list(per_cluster_scores(point,
cluster_ids,
tree,
max_lambda_dict,
point_dict
).values()))
result /= result.sum()
return result
cluster_ids = tree._select_clusters()
raw_tree = tree._raw_tree
all_possible_clusters = np.arange(data.shape[0], raw_tree['parent'].max() + 1).astype(np.float64)
max_lambda_dict = {c:max_lambda_val(c, raw_tree) for c in all_possible_clusters}
point_dict = {c:set(points_in_cluster(c, raw_tree)) for c in all_possible_clusters}
colors = np.empty((data.shape[0], 3))
for x in range(data.shape[0]):
membership_vector = outlier_membership_vector(x, cluster_ids, raw_tree,
max_lambda_dict, point_dict, False)
color = np.argmax(membership_vector)
saturation = membership_vector[color]
colors[x] = sns.desaturate(pal[color], saturation)
plt.scatter(data.T[0], data.T[1], c=colors, **plot_kwds);
Middle Way
To combine distance and membership approaches we can: - view membership vectors as probability distributions - combining them can be achieved via Bayes’ rule
def combined_membership_vector(point, data, tree, exemplar_dict, cluster_ids,
max_lambda_dict, point_dict, softmax=False):
raw_tree = tree._raw_tree
dist_vec = dist_membership_vector(point, exemplar_dict, data, softmax)
outl_vec = outlier_membership_vector(point, cluster_ids, raw_tree,
max_lambda_dict, point_dict, softmax)
result = dist_vec * outl_vec
result /= result.sum()
return result
colors = np.empty((data.shape[0], 3))
for x in range(data.shape[0]):
membership_vector = combined_membership_vector(x, data, tree, exemplar_dict, cluster_ids,
max_lambda_dict, point_dict, False)
color = np.argmax(membership_vector)
saturation = membership_vector[color]
colors[x] = sns.desaturate(pal[color], saturation)
plt.scatter(data.T[0], data.T[1], c=colors, **plot_kwds);C:\Users\Jonathan\AppData\Local\Temp\ipykernel_16588\1075745609.py:16: RuntimeWarning: divide by zero encountered in divide
result = 1./dist_vector(point, exemplar_dict, data)

Converting a Conditional Probability
The previous computation was the probability vector that a point is in each cluster, given that the point is in a cluster. To convert the conditional probability to a joint one, we need to: - multiply it by the probability that that there is a cluster to which the point belongs - this can be estimated from the merge height and comparing it with the max density - this should result in a number between 0 and 1
- plotting can show the $ $ cluster and the corresponding color
def prob_in_some_cluster(point, tree, cluster_ids, point_dict, max_lambda_dict):
heights = []
for cluster in cluster_ids:
heights.append(merge_height(point, cluster, tree._raw_tree, point_dict))
height = max(heights)
nearest_cluster = cluster_ids[np.argmax(heights)]
max_lambda = max_lambda_dict[nearest_cluster]
return height / max_lambda
colors = np.empty((data.shape[0], 3))
for x in range(data.shape[0]):
membership_vector = combined_membership_vector(x, data, tree, exemplar_dict, cluster_ids,
max_lambda_dict, point_dict, False)
membership_vector *= prob_in_some_cluster(x, tree, cluster_ids, point_dict, max_lambda_dict)
color = np.argmax(membership_vector)
saturation = membership_vector[color]
colors[x] = sns.desaturate(pal[color], saturation)
plt.scatter(data.T[0], data.T[1], c=colors, **plot_kwds);C:\Users\Jonathan\AppData\Local\Temp\ipykernel_16588\1075745609.py:16: RuntimeWarning: divide by zero encountered in divide
result = 1./dist_vector(point, exemplar_dict, data)

Soft Clustering in Practice
What are the general steps involved in performing a soft clustering analysis?
- dimensionality reduction as a prerequisite to 2D visualization:
- PCA
- TSNE
- UMAP
- visualization of raw data
- soft clustering via HDBSCAN
- visualization of cluster membership
- hard clustering color coding
- soft clustering color saturation
- quantitative analysis of cluster membership probabilities
from sklearn import datasets
from sklearn.manifold import TSNE
import matplotlib.pyplot as plt
import seaborn as sns
import numpy as np
import hdbscan
# loading raw data, 8x8 gray scale of handwritten digits, plotting low dimensional projection
digits = datasets.load_digits()
data = digits.data
projection = TSNE().fit_transform(data)
plt.scatter(*projection.T, **plot_kwds)
# setting up a clustering object, fitting and clustering the data
clusterer = hdbscan.HDBSCAN(min_cluster_size=10, prediction_data=True).fit(data)
color_palette = sns.color_palette('Paired', 12)
cluster_colors = [color_palette[x] if x >= 0
else (0.5, 0.5, 0.5)
for x in clusterer.labels_]
cluster_member_colors = [sns.desaturate(x, p) for x, p in
zip(cluster_colors, clusterer.probabilities_)]
plt.scatter(*projection.T, s=50, linewidth=0, c=cluster_member_colors, alpha=0.25)
Some of the data is noisy, so we can examine the noisy ones more through soft clustering.
soft_clusters = hdbscan.all_points_membership_vectors(clusterer)
color_palette = sns.color_palette('Paired', 12)
cluster_colors = [color_palette[np.argmax(x)]
for x in soft_clusters]
plt.scatter(*projection.T, s=50, linewidth=0, c=cluster_colors, alpha=0.25)
We can show uncertainty by coupling probability with desaturation of color. The higher the probability, the more saturated or pure the color. The lower the probability, the more gray it will be. Here we can see that desaturation is a harsh treatment with lots of gray, and that a lower limit may be more visually meaningful.
color_palette = sns.color_palette('Paired', 12)
cluster_colors = [sns.desaturate(color_palette[np.argmax(x)], np.max(x))
for x in soft_clusters]
plt.scatter(*projection.T, s=50, linewidth=0, c=cluster_colors, alpha=0.25)
One question to investigate is what points have high likelihoods for two clusters and low for the others. It’s worthwhile to note that the probabilities are joint ones, and that points have a probability of not being in a cluster.
def top_two_probs_diff(probs):
sorted_probs = np.sort(probs)
return sorted_probs[-1] - sorted_probs[-2]
# Compute the differences between the top two probabilities
diffs = np.array([top_two_probs_diff(x) for x in soft_clusters])
# Select out the indices that have a small difference, and a larger total probability, extract from tuple form
mixed_points = np.where((diffs < 0.001) & (np.sum(soft_clusters, axis=1) > 0.5))[0]
colors = [(0.75, 0.1, 0.1) if x in mixed_points
else (0.5, 0.5, 0.5) for x in range(data.shape[0])]
plt.scatter(*projection.T, s=50, linewidth=0, c=colors, alpha=0.5)
fig = plt.figure()
for i, image in enumerate(digits.images[mixed_points][:16]):
ax = fig.add_subplot(4,4,i+1)
ax.imshow(image)
plt.tight_layout()
plt.hist(np.sum(soft_clusters, axis=1))(array([ 36., 47., 92., 171., 332., 229., 181., 297., 153., 259.]),
array([0.58730357, 0.62857321, 0.66984285, 0.7111125 , 0.75238214,
0.79365178, 0.83492143, 0.87619107, 0.91746071, 0.95873036,
1. ]),
<BarContainer object of 10 artists>)

K-means Clustering
Description
Given a set of observations \((x_1, x_2, ..., x_n)\) in a \(d\)-dimensional vector space, k-means clustering partitions the \(n\) observations into \(k\) sets, \(\textbf{S} = \{S_1, S_2, ... S_k\}\). The goal is to minimize the variance within each cluster:
\[\underset{\textbf{S}}{\arg\min} \sum^k_{i=1} \sum_{x\in S_i} ||x - \mu_i||^2 = \underset{\textbf{S}}{\arg\min}\sum^k_{i=1}|S_i|\text{Var}S_i\]
where \(\mu_i\) is the mean or centroid of the points in \(S_i\).
Steps
Initialization
First choose \(k\) and initialize randomly chosen \(k\) centroids at \(t = 0\), the initial step.
Assignment
Assign each observation to the cluster with the nearest mean, in terms of Euclidean distance. Each set is thus:
\[S^{(t)}_i = \big\{x_p : ||x_p - m_i^{(t)}||^2 \le ||x_p - m_j^{(t)}||^2 \forall j, 1 \le j \le k \big\}\]
Update
After assignment, update to find the new means for the next iteration, \(t + 1\):
\[ m_i^{t + 1} = \frac{1}{|S_i^{(t)}|}\sum_{x_j \in S_i^{(t)}}x_j\]
The number of computations scales with the number of samples \(n\), the number of centers \(k\), the number of dimensions \(d\) and the number of iterations \(i\), for \(O(nkdi)\)
Creating the Dataset
# import statements
from sklearn.datasets import make_blobs
import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns
sns.set('notebook')
# Generate and plot data with predetermined model and classes
X, y = make_blobs(n_samples=200, n_features=2, centers=4,
cluster_std=1.6, random_state=50)
plt.scatter(X[:,0], X[:,1], c=y, cmap='viridis')
plt.xlim(-15,15)
plt.ylim(-15,15)
plt.show()
Y = np.zeros((200, 2))
Y[:,0] = y
Yarray([[2., 0.],
[0., 0.],
[3., 0.],
[3., 0.],
[1., 0.],
[0., 0.],
[2., 0.],
[0., 0.],
[0., 0.],
[1., 0.],
[1., 0.],
[2., 0.],
[1., 0.],
[1., 0.],
[1., 0.],
[3., 0.],
[2., 0.],
[2., 0.],
[1., 0.],
[1., 0.],
[1., 0.],
[3., 0.],
[2., 0.],
[2., 0.],
[2., 0.],
[3., 0.],
[2., 0.],
[1., 0.],
[3., 0.],
[2., 0.],
[2., 0.],
[1., 0.],
[1., 0.],
[0., 0.],
[0., 0.],
[1., 0.],
[1., 0.],
[1., 0.],
[3., 0.],
[3., 0.],
[2., 0.],
[1., 0.],
[3., 0.],
[3., 0.],
[0., 0.],
[0., 0.],
[2., 0.],
[3., 0.],
[3., 0.],
[2., 0.],
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[1., 0.],
[3., 0.],
[2., 0.],
[3., 0.],
[3., 0.],
[0., 0.],
[1., 0.],
[1., 0.],
[3., 0.],
[0., 0.],
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[2., 0.],
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[0., 0.],
[3., 0.],
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[2., 0.],
[0., 0.],
[1., 0.],
[2., 0.],
[0., 0.],
[3., 0.],
[3., 0.],
[1., 0.],
[3., 0.],
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[3., 0.],
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[1., 0.],
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[3., 0.],
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[2., 0.],
[3., 0.],
[0., 0.],
[2., 0.],
[2., 0.],
[2., 0.],
[2., 0.],
[0., 0.],
[2., 0.],
[2., 0.],
[3., 0.],
[2., 0.],
[0., 0.],
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[3., 0.],
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[2., 0.],
[3., 0.],
[2., 0.],
[1., 0.],
[1., 0.],
[2., 0.],
[0., 0.],
[3., 0.],
[0., 0.],
[1., 0.],
[2., 0.],
[1., 0.],
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[0., 0.],
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[3., 0.],
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[3., 0.],
[3., 0.],
[1., 0.],
[3., 0.],
[0., 0.],
[2., 0.],
[1., 0.],
[3., 0.],
[1., 0.],
[2., 0.]])
from sklearn.cluster import KMeansn_clusters=4
# Fit and predict clusters
kmeans = KMeans(n_clusters=n_clusters)
kmeans.fit(X)
print(kmeans.cluster_centers_)
y_km = kmeans.fit_predict(X)
# Plot predictions, one class at a time
for i, c in enumerate(['red', 'black', 'blue', 'cyan']):
plt.scatter(X[y_km ==i,0], X[y_km == i,1], s=100, c=c)
plt.show()[[-5.56465793 -2.34988939]
[-1.92101646 5.21673484]
[ 0.05161133 -5.35489826]
[-2.40167949 10.17352695]]

Hierarchical Clustering
Hierarchical clustering involves starting with treating each observation as a set, and then at the following step, creating a new cluster from the two “nearest” clusters, according to the defined distance metric. There are multiple ways the idea of a distance metric can extend to clusters. Some common choices include the average, closest members, or farthest nembers.
# import hierarchical clustering libraries
from sklearn.cluster import AgglomerativeClustering
from matplotlib import pyplot as plt
import seaborn as sns
from scipy.cluster.hierarchy import dendrogram, linkage, to_tree
%load_ext autoreload
%autoreload 2
sns.set()
data = [[i] for i in [-30, 4, 1, 2, 5, 6, 10, 50, 75, 100]]def llf(id):
return str(id)Z = linkage(data, 'ward')
fig = plt.figure(figsize=(20, 10))
dn_truncated = dendrogram(Z, orientation='right', p=4, truncate_mode='lastp', leaf_label_func=llf)
dn = dendrogram(Z, orientation='right', p=4, truncate_mode=None, leaf_label_func=llf)
dn_truncated{'icoord': [[5.0, 5.0, 15.0, 15.0],
[25.0, 25.0, 35.0, 35.0],
[10.0, 10.0, 30.0, 30.0]],
'dcoord': [[0.0, 43.30127018922193, 43.30127018922193, 0.0],
[0.0, 45.389321169086415, 45.389321169086415, 0.0],
[43.30127018922193,
154.2898015331631,
154.2898015331631,
45.389321169086415]],
'ivl': ['9', '15', '0', '14'],
'leaves': [9, 15, 0, 14],
'color_list': ['C1', 'C2', 'C0'],
'leaves_color_list': ['C1', 'C1', 'C2', 'C2'],
'traversal': [18, 16, 9, 15, 17, 0, 14]}
dn['icoord'][[15.0, 15.0, 25.0, 25.0],
[5.0, 5.0, 20.0, 20.0],
[55.0, 55.0, 65.0, 65.0],
[85.0, 85.0, 95.0, 95.0],
[75.0, 75.0, 90.0, 90.0],
[60.0, 60.0, 82.5, 82.5],
[45.0, 45.0, 71.25, 71.25],
[35.0, 35.0, 58.125, 58.125],
[12.5, 12.5, 46.5625, 46.5625]]
dn['dcoord'][[0.0, 25.0, 25.0, 0.0],
[0.0, 43.30127018922193, 43.30127018922193, 25.0],
[0.0, 1.0, 1.0, 0.0],
[0.0, 1.0, 1.0, 0.0],
[0.0, 1.7320508075688772, 1.7320508075688772, 1.0],
[1.0, 5.422176684690383, 5.422176684690383, 1.7320508075688772],
[0.0, 8.262364471909155, 8.262364471909155, 5.422176684690383],
[0.0, 45.389321169086415, 45.389321169086415, 8.262364471909155],
[43.30127018922193, 154.2898015331631, 154.2898015331631, 45.389321169086415]]
dn_truncated['icoord'][[5.0, 5.0, 15.0, 15.0], [25.0, 25.0, 35.0, 35.0], [10.0, 10.0, 30.0, 30.0]]
dn_truncated['dcoord'][[0.0, 43.30127018922193, 43.30127018922193, 0.0],
[0.0, 45.389321169086415, 45.389321169086415, 0.0],
[43.30127018922193, 154.2898015331631, 154.2898015331631, 45.389321169086415]]
k = 4
for i in zip(dn['icoord'][k], dn['dcoord'][k]):
print(i)(75.0, 0.0)
(75.0, 1.7320508075688772)
(90.0, 1.7320508075688772)
(90.0, 1.0)
Linkage Matrix
Full documentation for Scipy’s implementation is found here. The original observations are treated as \(n\) clusters. There are \(n-1\) new clusters created beyond these total, with indices ranging from \(n+1\) to \(2n-1\), ordered by the inter-cluster distance. Each of thse new clusters is made from two other clusters. The \(n-1 \times 4\) linkage matrix \(Z\) encodes information about the new clusters in the following manner: Each row \(i\) (keeping in mind zero indexing) describes the \((i+1)\)th cluster by listing the indices of the two source clusters, the inter-cluster distance between these two clusters, and finally, the number of members of the new cluster.
Zarray([[ 2. , 3. , 1. , 2. ],
[ 1. , 4. , 1. , 2. ],
[ 5. , 11. , 1.73205081, 3. ],
[ 10. , 12. , 5.42217668, 5. ],
[ 6. , 13. , 8.26236447, 6. ],
[ 7. , 8. , 25. , 2. ],
[ 9. , 15. , 43.30127019, 3. ],
[ 0. , 14. , 45.38932117, 7. ],
[ 16. , 17. , 154.28980153, 10. ]])
plt.show()Converting the Linkage Matrix to Tree
to_tree(Z)
rootnode, nodelist = to_tree(Z, rd=True)
rootnode<__main__.ClusterNode at 0x29427b86370>
rootnode.get_right().pre_order()[0, 6, 2, 3, 5, 1, 4]
len(nodelist)19
nodelist[4].pre_order()[4]
Visualization with Plotly
p =6np.array(data)array([[-30],
[ 4],
[ 1],
[ 2],
[ 5],
[ 6],
[ 10],
[ 50],
[ 75],
[100]])
n = len(data)len(range(2*n-p + 1, 2*n))5
traversal = dn_truncated['traversal']traversal.reverse()len(traversal)7
traversal[p:][14]
# import plotly.figure_factory as ff
import numpy as np
np.random.seed(1)
n = len(data)
p = 4
fig = create_dendrogram(np.array(data), hovertext=traversal[p:],
labels=data, orientation='left', p=p, truncate_mode='lastp', leaf_label_func=llf)
fig.update_layout(width=800, height=500)
fig.show()nodelist[13].pre_order()[2, 3, 5, 1, 4]
dn_truncated{'icoord': [[15.0, 15.0, 25.0, 25.0],
[5.0, 5.0, 20.0, 20.0],
[45.0, 45.0, 55.0, 55.0],
[35.0, 35.0, 50.0, 50.0],
[12.5, 12.5, 42.5, 42.5]],
'dcoord': [[0.0, 25.0, 25.0, 0.0],
[0.0, 43.30127018922193, 43.30127018922193, 25.0],
[0.0, 8.262364471909155, 8.262364471909155, 0.0],
[0.0, 45.389321169086415, 45.389321169086415, 8.262364471909155],
[43.30127018922193,
154.2898015331631,
154.2898015331631,
45.389321169086415]],
'ivl': ['9', '7', '8', '0', '6', '13'],
'leaves': [9, 7, 8, 0, 6, 13],
'color_list': ['C1', 'C1', 'C2', 'C2', 'C0'],
'leaves_color_list': ['C1', 'C1', 'C1', 'C2', 'C2', 'C2']}
Zarray([[ 2. , 3. , 1. , 2. ],
[ 1. , 4. , 1. , 2. ],
[ 5. , 11. , 1.73205081, 3. ],
[ 10. , 12. , 5.42217668, 5. ],
[ 6. , 13. , 8.26236447, 6. ],
[ 7. , 8. , 25. , 2. ],
[ 9. , 15. , 43.30127019, 3. ],
[ 0. , 14. , 45.38932117, 7. ],
[ 16. , 17. , 154.28980153, 10. ]])
dn{'icoord': [[15.0, 15.0, 25.0, 25.0],
[5.0, 5.0, 20.0, 20.0],
[55.0, 55.0, 65.0, 65.0],
[85.0, 85.0, 95.0, 95.0],
[75.0, 75.0, 90.0, 90.0],
[60.0, 60.0, 82.5, 82.5],
[45.0, 45.0, 71.25, 71.25],
[35.0, 35.0, 58.125, 58.125],
[12.5, 12.5, 46.5625, 46.5625]],
'dcoord': [[0.0, 25.0, 25.0, 0.0],
[0.0, 43.30127018922193, 43.30127018922193, 25.0],
[0.0, 1.0, 1.0, 0.0],
[0.0, 1.0, 1.0, 0.0],
[0.0, 1.7320508075688772, 1.7320508075688772, 1.0],
[1.0, 5.422176684690383, 5.422176684690383, 1.7320508075688772],
[0.0, 8.262364471909155, 8.262364471909155, 5.422176684690383],
[0.0, 45.389321169086415, 45.389321169086415, 8.262364471909155],
[43.30127018922193,
154.2898015331631,
154.2898015331631,
45.389321169086415]],
'ivl': ['9', '7', '8', '0', '6', '2', '3', '5', '1', '4'],
'leaves': [9, 7, 8, 0, 6, 2, 3, 5, 1, 4],
'color_list': ['C1', 'C1', 'C2', 'C2', 'C2', 'C2', 'C2', 'C2', 'C0'],
'leaves_color_list': ['C1',
'C1',
'C1',
'C2',
'C2',
'C2',
'C2',
'C2',
'C2',
'C2'],
'traversal': [18,
16,
9,
15,
7,
8,
17,
0,
14,
6,
13,
10,
2,
3,
12,
5,
11,
1,
4]}
dn_truncated{'icoord': [[15.0, 15.0, 25.0, 25.0],
[5.0, 5.0, 20.0, 20.0],
[45.0, 45.0, 55.0, 55.0],
[35.0, 35.0, 50.0, 50.0],
[12.5, 12.5, 42.5, 42.5]],
'dcoord': [[0.0, 25.0, 25.0, 0.0],
[0.0, 43.30127018922193, 43.30127018922193, 25.0],
[0.0, 8.262364471909155, 8.262364471909155, 0.0],
[0.0, 45.389321169086415, 45.389321169086415, 8.262364471909155],
[43.30127018922193,
154.2898015331631,
154.2898015331631,
45.389321169086415]],
'ivl': ['9', '7', '8', '0', '6', '(5)'],
'leaves': [9, 7, 8, 0, 6, 13],
'color_list': ['C1', 'C1', 'C2', 'C2', 'C0'],
'leaves_color_list': ['C1', 'C1', 'C1', 'C2', 'C2', 'C2']}
i = 0
print(dn_truncated['dcoord'][i])
print(dn_truncated['icoord'][i])[0.0, 25.0, 25.0, 0.0]
[15.0, 15.0, 25.0, 25.0]
dn_truncated['dcoord'][[0.0, 25.0, 25.0, 0.0],
[0.0, 43.30127018922193, 43.30127018922193, 25.0],
[0.0, 8.262364471909155, 8.262364471909155, 0.0],
[0.0, 45.389321169086415, 45.389321169086415, 8.262364471909155],
[43.30127018922193, 154.2898015331631, 154.2898015331631, 45.389321169086415]]
dn_truncated['icoord'][[15.0, 15.0, 25.0, 25.0],
[5.0, 5.0, 20.0, 20.0],
[45.0, 45.0, 55.0, 55.0],
[35.0, 35.0, 50.0, 50.0],
[12.5, 12.5, 42.5, 42.5]]
nodelist[].pre_order()[4]
nodelist[9].iddata[2][1]
#from plotly.figure_factory import create_dendrogram
import numpy as np
import pandas as pd
Index= ['A','B','C','D','E','F','G','H','I','J']
df = pd.DataFrame(abs(np.random.randn(10, 10)), index=Index)
fig = create_dendrogram(df, labels=Index, orientation='left', hovertext=list(range(19)))
fig.show()Implementing Cluster Order in Scipy
Modification of _dendrogram_calculate_info function
_dendrogram_calculate_infonow has a list to track the cluster indices during construction of the dendrogram
"""
Hierarchical clustering (:mod:`scipy.cluster.hierarchy`)
========================================================
.. currentmodule:: scipy.cluster.hierarchy
These functions cut hierarchical clusterings into flat clusterings
or find the roots of the forest formed by a cut by providing the flat
cluster ids of each observation.
.. autosummary::
:toctree: generated/
fcluster
fclusterdata
leaders
These are routines for agglomerative clustering.
.. autosummary::
:toctree: generated/
linkage
single
complete
average
weighted
centroid
median
ward
These routines compute statistics on hierarchies.
.. autosummary::
:toctree: generated/
cophenet
from_mlab_linkage
inconsistent
maxinconsts
maxdists
maxRstat
to_mlab_linkage
Routines for visualizing flat clusters.
.. autosummary::
:toctree: generated/
dendrogram
These are data structures and routines for representing hierarchies as
tree objects.
.. autosummary::
:toctree: generated/
ClusterNode
leaves_list
to_tree
cut_tree
optimal_leaf_ordering
These are predicates for checking the validity of linkage and
inconsistency matrices as well as for checking isomorphism of two
flat cluster assignments.
.. autosummary::
:toctree: generated/
is_valid_im
is_valid_linkage
is_isomorphic
is_monotonic
correspond
num_obs_linkage
Utility routines for plotting:
.. autosummary::
:toctree: generated/
set_link_color_palette
Utility classes:
.. autosummary::
:toctree: generated/
DisjointSet -- data structure for incremental connectivity queries
"""
# Copyright (C) Damian Eads, 2007-2008. New BSD License.
# hierarchy.py (derived from cluster.py, http://scipy-cluster.googlecode.com)
#
# Author: Damian Eads
# Date: September 22, 2007
#
# Copyright (c) 2007, 2008, Damian Eads
#
# All rights reserved.
#
# Redistribution and use in source and binary forms, with or without
# modification, are permitted provided that the following conditions
# are met:
# - Redistributions of source code must retain the above
# copyright notice, this list of conditions and the
# following disclaimer.
# - Redistributions in binary form must reproduce the above copyright
# notice, this list of conditions and the following disclaimer
# in the documentation and/or other materials provided with the
# distribution.
# - Neither the name of the author nor the names of its
# contributors may be used to endorse or promote products derived
# from this software without specific prior written permission.
#
# THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS
# "AS IS" AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT
# LIMITED TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR
# A PARTICULAR PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT
# OWNER OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL,
# SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT
# LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE,
# DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY
# THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT
# (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE
# OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.
import warnings
import bisect
from collections import deque
import numpy as np
from scipy.cluster import _hierarchy, _optimal_leaf_ordering
import scipy.spatial.distance as distance
from scipy._lib._disjoint_set import DisjointSet
_LINKAGE_METHODS = {'single': 0, 'complete': 1, 'average': 2, 'centroid': 3,
'median': 4, 'ward': 5, 'weighted': 6}
_EUCLIDEAN_METHODS = ('centroid', 'median', 'ward')
__all__ = ['ClusterNode', 'DisjointSet', 'average', 'centroid', 'complete',
'cophenet', 'correspond', 'cut_tree', 'dendrogram', 'fcluster',
'fclusterdata', 'from_mlab_linkage', 'inconsistent',
'is_isomorphic', 'is_monotonic', 'is_valid_im', 'is_valid_linkage',
'leaders', 'leaves_list', 'linkage', 'maxRstat', 'maxdists',
'maxinconsts', 'median', 'num_obs_linkage', 'optimal_leaf_ordering',
'set_link_color_palette', 'single', 'to_mlab_linkage', 'to_tree',
'ward', 'weighted', 'distance']
class ClusterWarning(UserWarning):
pass
def _warning(s):
warnings.warn('scipy.cluster: %s' % s, ClusterWarning, stacklevel=3)
def _copy_array_if_base_present(a):
"""
Copy the array if its base points to a parent array.
"""
if a.base is not None:
return a.copy()
elif np.issubsctype(a, np.float32):
return np.array(a, dtype=np.double)
else:
return a
def _copy_arrays_if_base_present(T):
"""
Accept a tuple of arrays T. Copies the array T[i] if its base array
points to an actual array. Otherwise, the reference is just copied.
This is useful if the arrays are being passed to a C function that
does not do proper striding.
"""
l = [_copy_array_if_base_present(a) for a in T]
return l
def _randdm(pnts):
"""
Generate a random distance matrix stored in condensed form.
Parameters
----------
pnts : int
The number of points in the distance matrix. Has to be at least 2.
Returns
-------
D : ndarray
A ``pnts * (pnts - 1) / 2`` sized vector is returned.
"""
if pnts >= 2:
D = np.random.rand(pnts * (pnts - 1) / 2)
else:
raise ValueError("The number of points in the distance matrix "
"must be at least 2.")
return D
def single(y):
"""
Perform single/min/nearest linkage on the condensed distance matrix ``y``.
Parameters
----------
y : ndarray
The upper triangular of the distance matrix. The result of
``pdist`` is returned in this form.
Returns
-------
Z : ndarray
The linkage matrix.
See Also
--------
linkage: for advanced creation of hierarchical clusterings.
scipy.spatial.distance.pdist : pairwise distance metrics
Examples
--------
>>> from scipy.cluster.hierarchy import single, fcluster
>>> from scipy.spatial.distance import pdist
First, we need a toy dataset to play with::
x x x x
x x
x x
x x x x
>>> X = [[0, 0], [0, 1], [1, 0],
... [0, 4], [0, 3], [1, 4],
... [4, 0], [3, 0], [4, 1],
... [4, 4], [3, 4], [4, 3]]
Then, we get a condensed distance matrix from this dataset:
>>> y = pdist(X)
Finally, we can perform the clustering:
>>> Z = single(y)
>>> Z
array([[ 0., 1., 1., 2.],
[ 2., 12., 1., 3.],
[ 3., 4., 1., 2.],
[ 5., 14., 1., 3.],
[ 6., 7., 1., 2.],
[ 8., 16., 1., 3.],
[ 9., 10., 1., 2.],
[11., 18., 1., 3.],
[13., 15., 2., 6.],
[17., 20., 2., 9.],
[19., 21., 2., 12.]])
The linkage matrix ``Z`` represents a dendrogram - see
`scipy.cluster.hierarchy.linkage` for a detailed explanation of its
contents.
We can use `scipy.cluster.hierarchy.fcluster` to see to which cluster
each initial point would belong given a distance threshold:
>>> fcluster(Z, 0.9, criterion='distance')
array([ 7, 8, 9, 10, 11, 12, 4, 5, 6, 1, 2, 3], dtype=int32)
>>> fcluster(Z, 1, criterion='distance')
array([3, 3, 3, 4, 4, 4, 2, 2, 2, 1, 1, 1], dtype=int32)
>>> fcluster(Z, 2, criterion='distance')
array([1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], dtype=int32)
Also, `scipy.cluster.hierarchy.dendrogram` can be used to generate a
plot of the dendrogram.
"""
return linkage(y, method='single', metric='euclidean')
def complete(y):
"""
Perform complete/max/farthest point linkage on a condensed distance matrix.
Parameters
----------
y : ndarray
The upper triangular of the distance matrix. The result of
``pdist`` is returned in this form.
Returns
-------
Z : ndarray
A linkage matrix containing the hierarchical clustering. See
the `linkage` function documentation for more information
on its structure.
See Also
--------
linkage: for advanced creation of hierarchical clusterings.
scipy.spatial.distance.pdist : pairwise distance metrics
Examples
--------
>>> from scipy.cluster.hierarchy import complete, fcluster
>>> from scipy.spatial.distance import pdist
First, we need a toy dataset to play with::
x x x x
x x
x x
x x x x
>>> X = [[0, 0], [0, 1], [1, 0],
... [0, 4], [0, 3], [1, 4],
... [4, 0], [3, 0], [4, 1],
... [4, 4], [3, 4], [4, 3]]
Then, we get a condensed distance matrix from this dataset:
>>> y = pdist(X)
Finally, we can perform the clustering:
>>> Z = complete(y)
>>> Z
array([[ 0. , 1. , 1. , 2. ],
[ 3. , 4. , 1. , 2. ],
[ 6. , 7. , 1. , 2. ],
[ 9. , 10. , 1. , 2. ],
[ 2. , 12. , 1.41421356, 3. ],
[ 5. , 13. , 1.41421356, 3. ],
[ 8. , 14. , 1.41421356, 3. ],
[11. , 15. , 1.41421356, 3. ],
[16. , 17. , 4.12310563, 6. ],
[18. , 19. , 4.12310563, 6. ],
[20. , 21. , 5.65685425, 12. ]])
The linkage matrix ``Z`` represents a dendrogram - see
`scipy.cluster.hierarchy.linkage` for a detailed explanation of its
contents.
We can use `scipy.cluster.hierarchy.fcluster` to see to which cluster
each initial point would belong given a distance threshold:
>>> fcluster(Z, 0.9, criterion='distance')
array([ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], dtype=int32)
>>> fcluster(Z, 1.5, criterion='distance')
array([1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4], dtype=int32)
>>> fcluster(Z, 4.5, criterion='distance')
array([1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2], dtype=int32)
>>> fcluster(Z, 6, criterion='distance')
array([1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], dtype=int32)
Also, `scipy.cluster.hierarchy.dendrogram` can be used to generate a
plot of the dendrogram.
"""
return linkage(y, method='complete', metric='euclidean')
def average(y):
"""
Perform average/UPGMA linkage on a condensed distance matrix.
Parameters
----------
y : ndarray
The upper triangular of the distance matrix. The result of
``pdist`` is returned in this form.
Returns
-------
Z : ndarray
A linkage matrix containing the hierarchical clustering. See
`linkage` for more information on its structure.
See Also
--------
linkage: for advanced creation of hierarchical clusterings.
scipy.spatial.distance.pdist : pairwise distance metrics
Examples
--------
>>> from scipy.cluster.hierarchy import average, fcluster
>>> from scipy.spatial.distance import pdist
First, we need a toy dataset to play with::
x x x x
x x
x x
x x x x
>>> X = [[0, 0], [0, 1], [1, 0],
... [0, 4], [0, 3], [1, 4],
... [4, 0], [3, 0], [4, 1],
... [4, 4], [3, 4], [4, 3]]
Then, we get a condensed distance matrix from this dataset:
>>> y = pdist(X)
Finally, we can perform the clustering:
>>> Z = average(y)
>>> Z
array([[ 0. , 1. , 1. , 2. ],
[ 3. , 4. , 1. , 2. ],
[ 6. , 7. , 1. , 2. ],
[ 9. , 10. , 1. , 2. ],
[ 2. , 12. , 1.20710678, 3. ],
[ 5. , 13. , 1.20710678, 3. ],
[ 8. , 14. , 1.20710678, 3. ],
[11. , 15. , 1.20710678, 3. ],
[16. , 17. , 3.39675184, 6. ],
[18. , 19. , 3.39675184, 6. ],
[20. , 21. , 4.09206523, 12. ]])
The linkage matrix ``Z`` represents a dendrogram - see
`scipy.cluster.hierarchy.linkage` for a detailed explanation of its
contents.
We can use `scipy.cluster.hierarchy.fcluster` to see to which cluster
each initial point would belong given a distance threshold:
>>> fcluster(Z, 0.9, criterion='distance')
array([ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], dtype=int32)
>>> fcluster(Z, 1.5, criterion='distance')
array([1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4], dtype=int32)
>>> fcluster(Z, 4, criterion='distance')
array([1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2], dtype=int32)
>>> fcluster(Z, 6, criterion='distance')
array([1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], dtype=int32)
Also, `scipy.cluster.hierarchy.dendrogram` can be used to generate a
plot of the dendrogram.
"""
return linkage(y, method='average', metric='euclidean')
def weighted(y):
"""
Perform weighted/WPGMA linkage on the condensed distance matrix.
See `linkage` for more information on the return
structure and algorithm.
Parameters
----------
y : ndarray
The upper triangular of the distance matrix. The result of
``pdist`` is returned in this form.
Returns
-------
Z : ndarray
A linkage matrix containing the hierarchical clustering. See
`linkage` for more information on its structure.
See Also
--------
linkage : for advanced creation of hierarchical clusterings.
scipy.spatial.distance.pdist : pairwise distance metrics
Examples
--------
>>> from scipy.cluster.hierarchy import weighted, fcluster
>>> from scipy.spatial.distance import pdist
First, we need a toy dataset to play with::
x x x x
x x
x x
x x x x
>>> X = [[0, 0], [0, 1], [1, 0],
... [0, 4], [0, 3], [1, 4],
... [4, 0], [3, 0], [4, 1],
... [4, 4], [3, 4], [4, 3]]
Then, we get a condensed distance matrix from this dataset:
>>> y = pdist(X)
Finally, we can perform the clustering:
>>> Z = weighted(y)
>>> Z
array([[ 0. , 1. , 1. , 2. ],
[ 6. , 7. , 1. , 2. ],
[ 3. , 4. , 1. , 2. ],
[ 9. , 11. , 1. , 2. ],
[ 2. , 12. , 1.20710678, 3. ],
[ 8. , 13. , 1.20710678, 3. ],
[ 5. , 14. , 1.20710678, 3. ],
[10. , 15. , 1.20710678, 3. ],
[18. , 19. , 3.05595762, 6. ],
[16. , 17. , 3.32379407, 6. ],
[20. , 21. , 4.06357713, 12. ]])
The linkage matrix ``Z`` represents a dendrogram - see
`scipy.cluster.hierarchy.linkage` for a detailed explanation of its
contents.
We can use `scipy.cluster.hierarchy.fcluster` to see to which cluster
each initial point would belong given a distance threshold:
>>> fcluster(Z, 0.9, criterion='distance')
array([ 7, 8, 9, 1, 2, 3, 10, 11, 12, 4, 6, 5], dtype=int32)
>>> fcluster(Z, 1.5, criterion='distance')
array([3, 3, 3, 1, 1, 1, 4, 4, 4, 2, 2, 2], dtype=int32)
>>> fcluster(Z, 4, criterion='distance')
array([2, 2, 2, 1, 1, 1, 2, 2, 2, 1, 1, 1], dtype=int32)
>>> fcluster(Z, 6, criterion='distance')
array([1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], dtype=int32)
Also, `scipy.cluster.hierarchy.dendrogram` can be used to generate a
plot of the dendrogram.
"""
return linkage(y, method='weighted', metric='euclidean')
def centroid(y):
"""
Perform centroid/UPGMC linkage.
See `linkage` for more information on the input matrix,
return structure, and algorithm.
The following are common calling conventions:
1. ``Z = centroid(y)``
Performs centroid/UPGMC linkage on the condensed distance
matrix ``y``.
2. ``Z = centroid(X)``
Performs centroid/UPGMC linkage on the observation matrix ``X``
using Euclidean distance as the distance metric.
Parameters
----------
y : ndarray
A condensed distance matrix. A condensed
distance matrix is a flat array containing the upper
triangular of the distance matrix. This is the form that
``pdist`` returns. Alternatively, a collection of
m observation vectors in n dimensions may be passed as
an m by n array.
Returns
-------
Z : ndarray
A linkage matrix containing the hierarchical clustering. See
the `linkage` function documentation for more information
on its structure.
See Also
--------
linkage: for advanced creation of hierarchical clusterings.
scipy.spatial.distance.pdist : pairwise distance metrics
Examples
--------
>>> from scipy.cluster.hierarchy import centroid, fcluster
>>> from scipy.spatial.distance import pdist
First, we need a toy dataset to play with::
x x x x
x x
x x
x x x x
>>> X = [[0, 0], [0, 1], [1, 0],
... [0, 4], [0, 3], [1, 4],
... [4, 0], [3, 0], [4, 1],
... [4, 4], [3, 4], [4, 3]]
Then, we get a condensed distance matrix from this dataset:
>>> y = pdist(X)
Finally, we can perform the clustering:
>>> Z = centroid(y)
>>> Z
array([[ 0. , 1. , 1. , 2. ],
[ 3. , 4. , 1. , 2. ],
[ 9. , 10. , 1. , 2. ],
[ 6. , 7. , 1. , 2. ],
[ 2. , 12. , 1.11803399, 3. ],
[ 5. , 13. , 1.11803399, 3. ],
[ 8. , 15. , 1.11803399, 3. ],
[11. , 14. , 1.11803399, 3. ],
[18. , 19. , 3.33333333, 6. ],
[16. , 17. , 3.33333333, 6. ],
[20. , 21. , 3.33333333, 12. ]])
The linkage matrix ``Z`` represents a dendrogram - see
`scipy.cluster.hierarchy.linkage` for a detailed explanation of its
contents.
We can use `scipy.cluster.hierarchy.fcluster` to see to which cluster
each initial point would belong given a distance threshold:
>>> fcluster(Z, 0.9, criterion='distance')
array([ 7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6], dtype=int32)
>>> fcluster(Z, 1.1, criterion='distance')
array([5, 5, 6, 7, 7, 8, 1, 1, 2, 3, 3, 4], dtype=int32)
>>> fcluster(Z, 2, criterion='distance')
array([3, 3, 3, 4, 4, 4, 1, 1, 1, 2, 2, 2], dtype=int32)
>>> fcluster(Z, 4, criterion='distance')
array([1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], dtype=int32)
Also, `scipy.cluster.hierarchy.dendrogram` can be used to generate a
plot of the dendrogram.
"""
return linkage(y, method='centroid', metric='euclidean')
def median(y):
"""
Perform median/WPGMC linkage.
See `linkage` for more information on the return structure
and algorithm.
The following are common calling conventions:
1. ``Z = median(y)``
Performs median/WPGMC linkage on the condensed distance matrix
``y``. See ``linkage`` for more information on the return
structure and algorithm.
2. ``Z = median(X)``
Performs median/WPGMC linkage on the observation matrix ``X``
using Euclidean distance as the distance metric. See `linkage`
for more information on the return structure and algorithm.
Parameters
----------
y : ndarray
A condensed distance matrix. A condensed
distance matrix is a flat array containing the upper
triangular of the distance matrix. This is the form that
``pdist`` returns. Alternatively, a collection of
m observation vectors in n dimensions may be passed as
an m by n array.
Returns
-------
Z : ndarray
The hierarchical clustering encoded as a linkage matrix.
See Also
--------
linkage: for advanced creation of hierarchical clusterings.
scipy.spatial.distance.pdist : pairwise distance metrics
Examples
--------
>>> from scipy.cluster.hierarchy import median, fcluster
>>> from scipy.spatial.distance import pdist
First, we need a toy dataset to play with::
x x x x
x x
x x
x x x x
>>> X = [[0, 0], [0, 1], [1, 0],
... [0, 4], [0, 3], [1, 4],
... [4, 0], [3, 0], [4, 1],
... [4, 4], [3, 4], [4, 3]]
Then, we get a condensed distance matrix from this dataset:
>>> y = pdist(X)
Finally, we can perform the clustering:
>>> Z = median(y)
>>> Z
array([[ 0. , 1. , 1. , 2. ],
[ 3. , 4. , 1. , 2. ],
[ 9. , 10. , 1. , 2. ],
[ 6. , 7. , 1. , 2. ],
[ 2. , 12. , 1.11803399, 3. ],
[ 5. , 13. , 1.11803399, 3. ],
[ 8. , 15. , 1.11803399, 3. ],
[11. , 14. , 1.11803399, 3. ],
[18. , 19. , 3. , 6. ],
[16. , 17. , 3.5 , 6. ],
[20. , 21. , 3.25 , 12. ]])
The linkage matrix ``Z`` represents a dendrogram - see
`scipy.cluster.hierarchy.linkage` for a detailed explanation of its
contents.
We can use `scipy.cluster.hierarchy.fcluster` to see to which cluster
each initial point would belong given a distance threshold:
>>> fcluster(Z, 0.9, criterion='distance')
array([ 7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6], dtype=int32)
>>> fcluster(Z, 1.1, criterion='distance')
array([5, 5, 6, 7, 7, 8, 1, 1, 2, 3, 3, 4], dtype=int32)
>>> fcluster(Z, 2, criterion='distance')
array([3, 3, 3, 4, 4, 4, 1, 1, 1, 2, 2, 2], dtype=int32)
>>> fcluster(Z, 4, criterion='distance')
array([1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], dtype=int32)
Also, `scipy.cluster.hierarchy.dendrogram` can be used to generate a
plot of the dendrogram.
"""
return linkage(y, method='median', metric='euclidean')
def ward(y):
"""
Perform Ward's linkage on a condensed distance matrix.
See `linkage` for more information on the return structure
and algorithm.
The following are common calling conventions:
1. ``Z = ward(y)``
Performs Ward's linkage on the condensed distance matrix ``y``.
2. ``Z = ward(X)``
Performs Ward's linkage on the observation matrix ``X`` using
Euclidean distance as the distance metric.
Parameters
----------
y : ndarray
A condensed distance matrix. A condensed
distance matrix is a flat array containing the upper
triangular of the distance matrix. This is the form that
``pdist`` returns. Alternatively, a collection of
m observation vectors in n dimensions may be passed as
an m by n array.
Returns
-------
Z : ndarray
The hierarchical clustering encoded as a linkage matrix. See
`linkage` for more information on the return structure and
algorithm.
See Also
--------
linkage: for advanced creation of hierarchical clusterings.
scipy.spatial.distance.pdist : pairwise distance metrics
Examples
--------
>>> from scipy.cluster.hierarchy import ward, fcluster
>>> from scipy.spatial.distance import pdist
First, we need a toy dataset to play with::
x x x x
x x
x x
x x x x
>>> X = [[0, 0], [0, 1], [1, 0],
... [0, 4], [0, 3], [1, 4],
... [4, 0], [3, 0], [4, 1],
... [4, 4], [3, 4], [4, 3]]
Then, we get a condensed distance matrix from this dataset:
>>> y = pdist(X)
Finally, we can perform the clustering:
>>> Z = ward(y)
>>> Z
array([[ 0. , 1. , 1. , 2. ],
[ 3. , 4. , 1. , 2. ],
[ 6. , 7. , 1. , 2. ],
[ 9. , 10. , 1. , 2. ],
[ 2. , 12. , 1.29099445, 3. ],
[ 5. , 13. , 1.29099445, 3. ],
[ 8. , 14. , 1.29099445, 3. ],
[11. , 15. , 1.29099445, 3. ],
[16. , 17. , 5.77350269, 6. ],
[18. , 19. , 5.77350269, 6. ],
[20. , 21. , 8.16496581, 12. ]])
The linkage matrix ``Z`` represents a dendrogram - see
`scipy.cluster.hierarchy.linkage` for a detailed explanation of its
contents.
We can use `scipy.cluster.hierarchy.fcluster` to see to which cluster
each initial point would belong given a distance threshold:
>>> fcluster(Z, 0.9, criterion='distance')
array([ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], dtype=int32)
>>> fcluster(Z, 1.1, criterion='distance')
array([1, 1, 2, 3, 3, 4, 5, 5, 6, 7, 7, 8], dtype=int32)
>>> fcluster(Z, 3, criterion='distance')
array([1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4], dtype=int32)
>>> fcluster(Z, 9, criterion='distance')
array([1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], dtype=int32)
Also, `scipy.cluster.hierarchy.dendrogram` can be used to generate a
plot of the dendrogram.
"""
return linkage(y, method='ward', metric='euclidean')
def linkage(y, method='single', metric='euclidean', optimal_ordering=False):
"""
Perform hierarchical/agglomerative clustering.
The input y may be either a 1-D condensed distance matrix
or a 2-D array of observation vectors.
If y is a 1-D condensed distance matrix,
then y must be a :math:`\\binom{n}{2}` sized
vector, where n is the number of original observations paired
in the distance matrix. The behavior of this function is very
similar to the MATLAB linkage function.
A :math:`(n-1)` by 4 matrix ``Z`` is returned. At the
:math:`i`-th iteration, clusters with indices ``Z[i, 0]`` and
``Z[i, 1]`` are combined to form cluster :math:`n + i`. A
cluster with an index less than :math:`n` corresponds to one of
the :math:`n` original observations. The distance between
clusters ``Z[i, 0]`` and ``Z[i, 1]`` is given by ``Z[i, 2]``. The
fourth value ``Z[i, 3]`` represents the number of original
observations in the newly formed cluster.
The following linkage methods are used to compute the distance
:math:`d(s, t)` between two clusters :math:`s` and
:math:`t`. The algorithm begins with a forest of clusters that
have yet to be used in the hierarchy being formed. When two
clusters :math:`s` and :math:`t` from this forest are combined
into a single cluster :math:`u`, :math:`s` and :math:`t` are
removed from the forest, and :math:`u` is added to the
forest. When only one cluster remains in the forest, the algorithm
stops, and this cluster becomes the root.
A distance matrix is maintained at each iteration. The ``d[i,j]``
entry corresponds to the distance between cluster :math:`i` and
:math:`j` in the original forest.
At each iteration, the algorithm must update the distance matrix
to reflect the distance of the newly formed cluster u with the
remaining clusters in the forest.
Suppose there are :math:`|u|` original observations
:math:`u[0], \\ldots, u[|u|-1]` in cluster :math:`u` and
:math:`|v|` original objects :math:`v[0], \\ldots, v[|v|-1]` in
cluster :math:`v`. Recall, :math:`s` and :math:`t` are
combined to form cluster :math:`u`. Let :math:`v` be any
remaining cluster in the forest that is not :math:`u`.
The following are methods for calculating the distance between the
newly formed cluster :math:`u` and each :math:`v`.
* method='single' assigns
.. math::
d(u,v) = \\min(dist(u[i],v[j]))
for all points :math:`i` in cluster :math:`u` and
:math:`j` in cluster :math:`v`. This is also known as the
Nearest Point Algorithm.
* method='complete' assigns
.. math::
d(u, v) = \\max(dist(u[i],v[j]))
for all points :math:`i` in cluster u and :math:`j` in
cluster :math:`v`. This is also known by the Farthest Point
Algorithm or Voor Hees Algorithm.
* method='average' assigns
.. math::
d(u,v) = \\sum_{ij} \\frac{d(u[i], v[j])}
{(|u|*|v|)}
for all points :math:`i` and :math:`j` where :math:`|u|`
and :math:`|v|` are the cardinalities of clusters :math:`u`
and :math:`v`, respectively. This is also called the UPGMA
algorithm.
* method='weighted' assigns
.. math::
d(u,v) = (dist(s,v) + dist(t,v))/2
where cluster u was formed with cluster s and t and v
is a remaining cluster in the forest (also called WPGMA).
* method='centroid' assigns
.. math::
dist(s,t) = ||c_s-c_t||_2
where :math:`c_s` and :math:`c_t` are the centroids of
clusters :math:`s` and :math:`t`, respectively. When two
clusters :math:`s` and :math:`t` are combined into a new
cluster :math:`u`, the new centroid is computed over all the
original objects in clusters :math:`s` and :math:`t`. The
distance then becomes the Euclidean distance between the
centroid of :math:`u` and the centroid of a remaining cluster
:math:`v` in the forest. This is also known as the UPGMC
algorithm.
* method='median' assigns :math:`d(s,t)` like the ``centroid``
method. When two clusters :math:`s` and :math:`t` are combined
into a new cluster :math:`u`, the average of centroids s and t
give the new centroid :math:`u`. This is also known as the
WPGMC algorithm.
* method='ward' uses the Ward variance minimization algorithm.
The new entry :math:`d(u,v)` is computed as follows,
.. math::
d(u,v) = \\sqrt{\\frac{|v|+|s|}
{T}d(v,s)^2
+ \\frac{|v|+|t|}
{T}d(v,t)^2
- \\frac{|v|}
{T}d(s,t)^2}
where :math:`u` is the newly joined cluster consisting of
clusters :math:`s` and :math:`t`, :math:`v` is an unused
cluster in the forest, :math:`T=|v|+|s|+|t|`, and
:math:`|*|` is the cardinality of its argument. This is also
known as the incremental algorithm.
Warning: When the minimum distance pair in the forest is chosen, there
may be two or more pairs with the same minimum distance. This
implementation may choose a different minimum than the MATLAB
version.
Parameters
----------
y : ndarray
A condensed distance matrix. A condensed distance matrix
is a flat array containing the upper triangular of the distance matrix.
This is the form that ``pdist`` returns. Alternatively, a collection of
:math:`m` observation vectors in :math:`n` dimensions may be passed as
an :math:`m` by :math:`n` array. All elements of the condensed distance
matrix must be finite, i.e., no NaNs or infs.
method : str, optional
The linkage algorithm to use. See the ``Linkage Methods`` section below
for full descriptions.
metric : str or function, optional
The distance metric to use in the case that y is a collection of
observation vectors; ignored otherwise. See the ``pdist``
function for a list of valid distance metrics. A custom distance
function can also be used.
optimal_ordering : bool, optional
If True, the linkage matrix will be reordered so that the distance
between successive leaves is minimal. This results in a more intuitive
tree structure when the data are visualized. defaults to False, because
this algorithm can be slow, particularly on large datasets [2]_. See
also the `optimal_leaf_ordering` function.
.. versionadded:: 1.0.0
Returns
-------
Z : ndarray
The hierarchical clustering encoded as a linkage matrix.
Notes
-----
1. For method 'single', an optimized algorithm based on minimum spanning
tree is implemented. It has time complexity :math:`O(n^2)`.
For methods 'complete', 'average', 'weighted' and 'ward', an algorithm
called nearest-neighbors chain is implemented. It also has time
complexity :math:`O(n^2)`.
For other methods, a naive algorithm is implemented with :math:`O(n^3)`
time complexity.
All algorithms use :math:`O(n^2)` memory.
Refer to [1]_ for details about the algorithms.
2. Methods 'centroid', 'median', and 'ward' are correctly defined only if
Euclidean pairwise metric is used. If `y` is passed as precomputed
pairwise distances, then it is the user's responsibility to assure that
these distances are in fact Euclidean, otherwise the produced result
will be incorrect.
See Also
--------
scipy.spatial.distance.pdist : pairwise distance metrics
References
----------
.. [1] Daniel Mullner, "Modern hierarchical, agglomerative clustering
algorithms", :arXiv:`1109.2378v1`.
.. [2] Ziv Bar-Joseph, David K. Gifford, Tommi S. Jaakkola, "Fast optimal
leaf ordering for hierarchical clustering", 2001. Bioinformatics
:doi:`10.1093/bioinformatics/17.suppl_1.S22`
Examples
--------
>>> from scipy.cluster.hierarchy import dendrogram, linkage
>>> from matplotlib import pyplot as plt
>>> X = [[i] for i in [2, 8, 0, 4, 1, 9, 9, 0]]
>>> Z = linkage(X, 'ward')
>>> fig = plt.figure(figsize=(25, 10))
>>> dn = dendrogram(Z)
>>> Z = linkage(X, 'single')
>>> fig = plt.figure(figsize=(25, 10))
>>> dn = dendrogram(Z)
>>> plt.show()
"""
if method not in _LINKAGE_METHODS:
raise ValueError("Invalid method: {0}".format(method))
y = _convert_to_double(np.asarray(y, order='c'))
if y.ndim == 1:
distance.is_valid_y(y, throw=True, name='y')
[y] = _copy_arrays_if_base_present([y])
elif y.ndim == 2:
if method in _EUCLIDEAN_METHODS and metric != 'euclidean':
raise ValueError("Method '{0}' requires the distance metric "
"to be Euclidean".format(method))
if y.shape[0] == y.shape[1] and np.allclose(np.diag(y), 0):
if np.all(y >= 0) and np.allclose(y, y.T):
_warning('The symmetric non-negative hollow observation '
'matrix looks suspiciously like an uncondensed '
'distance matrix')
y = distance.pdist(y, metric)
else:
raise ValueError("`y` must be 1 or 2 dimensional.")
if not np.all(np.isfinite(y)):
raise ValueError("The condensed distance matrix must contain only "
"finite values.")
n = int(distance.num_obs_y(y))
method_code = _LINKAGE_METHODS[method]
if method == 'single':
result = _hierarchy.mst_single_linkage(y, n)
elif method in ['complete', 'average', 'weighted', 'ward']:
result = _hierarchy.nn_chain(y, n, method_code)
else:
result = _hierarchy.fast_linkage(y, n, method_code)
if optimal_ordering:
return optimal_leaf_ordering(result, y)
else:
return result
class ClusterNode(object):
"""
A tree node class for representing a cluster.
Leaf nodes correspond to original observations, while non-leaf nodes
correspond to non-singleton clusters.
The `to_tree` function converts a matrix returned by the linkage
function into an easy-to-use tree representation.
All parameter names are also attributes.
Parameters
----------
id : int
The node id.
left : ClusterNode instance, optional
The left child tree node.
right : ClusterNode instance, optional
The right child tree node.
dist : float, optional
Distance for this cluster in the linkage matrix.
count : int, optional
The number of samples in this cluster.
See Also
--------
to_tree : for converting a linkage matrix ``Z`` into a tree object.
"""
def __init__(self, id, left=None, right=None, dist=0, count=1):
if id < 0:
raise ValueError('The id must be non-negative.')
if dist < 0:
raise ValueError('The distance must be non-negative.')
if (left is None and right is not None) or \
(left is not None and right is None):
raise ValueError('Only full or proper binary trees are permitted.'
' This node has one child.')
if count < 1:
raise ValueError('A cluster must contain at least one original '
'observation.')
self.id = id
self.left = left
self.right = right
self.dist = dist
if self.left is None:
self.count = count
else:
self.count = left.count + right.count
def __lt__(self, node):
if not isinstance(node, ClusterNode):
raise ValueError("Can't compare ClusterNode "
"to type {}".format(type(node)))
return self.dist < node.dist
def __gt__(self, node):
if not isinstance(node, ClusterNode):
raise ValueError("Can't compare ClusterNode "
"to type {}".format(type(node)))
return self.dist > node.dist
def __eq__(self, node):
if not isinstance(node, ClusterNode):
raise ValueError("Can't compare ClusterNode "
"to type {}".format(type(node)))
return self.dist == node.dist
def get_id(self):
"""
The identifier of the target node.
For ``0 <= i < n``, `i` corresponds to original observation i.
For ``n <= i < 2n-1``, `i` corresponds to non-singleton cluster formed
at iteration ``i-n``.
Returns
-------
id : int
The identifier of the target node.
"""
return self.id
def get_count(self):
"""
The number of leaf nodes (original observations) belonging to
the cluster node nd. If the target node is a leaf, 1 is
returned.
Returns
-------
get_count : int
The number of leaf nodes below the target node.
"""
return self.count
def get_left(self):
"""
Return a reference to the left child tree object.
Returns
-------
left : ClusterNode
The left child of the target node. If the node is a leaf,
None is returned.
"""
return self.left
def get_right(self):
"""
Return a reference to the right child tree object.
Returns
-------
right : ClusterNode
The left child of the target node. If the node is a leaf,
None is returned.
"""
return self.right
def is_leaf(self):
"""
Return True if the target node is a leaf.
Returns
-------
leafness : bool
True if the target node is a leaf node.
"""
return self.left is None
def pre_order(self, func=(lambda x: x.id)):
"""
Perform pre-order traversal without recursive function calls.
When a leaf node is first encountered, ``func`` is called with
the leaf node as its argument, and its result is appended to
the list.
For example, the statement::
ids = root.pre_order(lambda x: x.id)
returns a list of the node ids corresponding to the leaf nodes
of the tree as they appear from left to right.
Parameters
----------
func : function
Applied to each leaf ClusterNode object in the pre-order traversal.
Given the ``i``-th leaf node in the pre-order traversal ``n[i]``,
the result of ``func(n[i])`` is stored in ``L[i]``. If not
provided, the index of the original observation to which the node
corresponds is used.
Returns
-------
L : list
The pre-order traversal.
"""
# Do a preorder traversal, caching the result. To avoid having to do
# recursion, we'll store the previous index we've visited in a vector.
n = self.count
curNode = [None] * (2 * n)
lvisited = set()
rvisited = set()
curNode[0] = self
k = 0
preorder = []
while k >= 0:
nd = curNode[k]
ndid = nd.id
if nd.is_leaf():
preorder.append(func(nd))
k = k - 1
else:
if ndid not in lvisited:
curNode[k + 1] = nd.left
lvisited.add(ndid)
k = k + 1
elif ndid not in rvisited:
curNode[k + 1] = nd.right
rvisited.add(ndid)
k = k + 1
# If we've visited the left and right of this non-leaf
# node already, go up in the tree.
else:
k = k - 1
return preorder
_cnode_bare = ClusterNode(0)
_cnode_type = type(ClusterNode)
def _order_cluster_tree(Z):
"""
Return clustering nodes in bottom-up order by distance.
Parameters
----------
Z : scipy.cluster.linkage array
The linkage matrix.
Returns
-------
nodes : list
A list of ClusterNode objects.
"""
q = deque()
tree = to_tree(Z)
q.append(tree)
nodes = []
while q:
node = q.popleft()
if not node.is_leaf():
bisect.insort_left(nodes, node)
q.append(node.get_right())
q.append(node.get_left())
return nodes
def cut_tree(Z, n_clusters=None, height=None):
"""
Given a linkage matrix Z, return the cut tree.
Parameters
----------
Z : scipy.cluster.linkage array
The linkage matrix.
n_clusters : array_like, optional
Number of clusters in the tree at the cut point.
height : array_like, optional
The height at which to cut the tree. Only possible for ultrametric
trees.
Returns
-------
cutree : array
An array indicating group membership at each agglomeration step. I.e.,
for a full cut tree, in the first column each data point is in its own
cluster. At the next step, two nodes are merged. Finally, all
singleton and non-singleton clusters are in one group. If `n_clusters`
or `height` are given, the columns correspond to the columns of
`n_clusters` or `height`.
Examples
--------
>>> from scipy import cluster
>>> np.random.seed(23)
>>> X = np.random.randn(50, 4)
>>> Z = cluster.hierarchy.ward(X)
>>> cutree = cluster.hierarchy.cut_tree(Z, n_clusters=[5, 10])
>>> cutree[:10]
array([[0, 0],
[1, 1],
[2, 2],
[3, 3],
[3, 4],
[2, 2],
[0, 0],
[1, 5],
[3, 6],
[4, 7]])
"""
nobs = num_obs_linkage(Z)
nodes = _order_cluster_tree(Z)
if height is not None and n_clusters is not None:
raise ValueError("At least one of either height or n_clusters "
"must be None")
elif height is None and n_clusters is None: # return the full cut tree
cols_idx = np.arange(nobs)
elif height is not None:
heights = np.array([x.dist for x in nodes])
cols_idx = np.searchsorted(heights, height)
else:
cols_idx = nobs - np.searchsorted(np.arange(nobs), n_clusters)
try:
n_cols = len(cols_idx)
except TypeError: # scalar
n_cols = 1
cols_idx = np.array([cols_idx])
groups = np.zeros((n_cols, nobs), dtype=int)
last_group = np.arange(nobs)
if 0 in cols_idx:
groups[0] = last_group
for i, node in enumerate(nodes):
idx = node.pre_order()
this_group = last_group.copy()
this_group[idx] = last_group[idx].min()
this_group[this_group > last_group[idx].max()] -= 1
if i + 1 in cols_idx:
groups[np.nonzero(i + 1 == cols_idx)[0]] = this_group
last_group = this_group
return groups.T
def to_tree(Z, rd=False):
"""
Convert a linkage matrix into an easy-to-use tree object.
The reference to the root `ClusterNode` object is returned (by default).
Each `ClusterNode` object has a ``left``, ``right``, ``dist``, ``id``,
and ``count`` attribute. The left and right attributes point to
ClusterNode objects that were combined to generate the cluster.
If both are None then the `ClusterNode` object is a leaf node, its count
must be 1, and its distance is meaningless but set to 0.
*Note: This function is provided for the convenience of the library
user. ClusterNodes are not used as input to any of the functions in this
library.*
Parameters
----------
Z : ndarray
The linkage matrix in proper form (see the `linkage`
function documentation).
rd : bool, optional
When False (default), a reference to the root `ClusterNode` object is
returned. Otherwise, a tuple ``(r, d)`` is returned. ``r`` is a
reference to the root node while ``d`` is a list of `ClusterNode`
objects - one per original entry in the linkage matrix plus entries
for all clustering steps. If a cluster id is
less than the number of samples ``n`` in the data that the linkage
matrix describes, then it corresponds to a singleton cluster (leaf
node).
See `linkage` for more information on the assignment of cluster ids
to clusters.
Returns
-------
tree : ClusterNode or tuple (ClusterNode, list of ClusterNode)
If ``rd`` is False, a `ClusterNode`.
If ``rd`` is True, a list of length ``2*n - 1``, with ``n`` the number
of samples. See the description of `rd` above for more details.
See Also
--------
linkage, is_valid_linkage, ClusterNode
Examples
--------
>>> from scipy.cluster import hierarchy
>>> x = np.random.rand(10).reshape(5, 2)
>>> Z = hierarchy.linkage(x)
>>> hierarchy.to_tree(Z)
<scipy.cluster.hierarchy.ClusterNode object at ...
>>> rootnode, nodelist = hierarchy.to_tree(Z, rd=True)
>>> rootnode
<scipy.cluster.hierarchy.ClusterNode object at ...
>>> len(nodelist)
9
"""
Z = np.asarray(Z, order='c')
is_valid_linkage(Z, throw=True, name='Z')
# Number of original objects is equal to the number of rows minus 1.
n = Z.shape[0] + 1
# Create a list full of None's to store the node objects
d = [None] * (n * 2 - 1)
# Create the nodes corresponding to the n original objects.
for i in range(0, n):
d[i] = ClusterNode(i)
nd = None
for i in range(0, n - 1):
fi = int(Z[i, 0])
fj = int(Z[i, 1])
if fi > i + n:
raise ValueError(('Corrupt matrix Z. Index to derivative cluster '
'is used before it is formed. See row %d, '
'column 0') % fi)
if fj > i + n:
raise ValueError(('Corrupt matrix Z. Index to derivative cluster '
'is used before it is formed. See row %d, '
'column 1') % fj)
nd = ClusterNode(i + n, d[fi], d[fj], Z[i, 2])
# ^ id ^ left ^ right ^ dist
if Z[i, 3] != nd.count:
raise ValueError(('Corrupt matrix Z. The count Z[%d,3] is '
'incorrect.') % i)
d[n + i] = nd
if rd:
return (nd, d)
else:
return nd
def optimal_leaf_ordering(Z, y, metric='euclidean'):
"""
Given a linkage matrix Z and distance, reorder the cut tree.
Parameters
----------
Z : ndarray
The hierarchical clustering encoded as a linkage matrix. See
`linkage` for more information on the return structure and
algorithm.
y : ndarray
The condensed distance matrix from which Z was generated.
Alternatively, a collection of m observation vectors in n
dimensions may be passed as an m by n array.
metric : str or function, optional
The distance metric to use in the case that y is a collection of
observation vectors; ignored otherwise. See the ``pdist``
function for a list of valid distance metrics. A custom distance
function can also be used.
Returns
-------
Z_ordered : ndarray
A copy of the linkage matrix Z, reordered to minimize the distance
between adjacent leaves.
Examples
--------
>>> from scipy.cluster import hierarchy
>>> np.random.seed(23)
>>> X = np.random.randn(10,10)
>>> Z = hierarchy.ward(X)
>>> hierarchy.leaves_list(Z)
array([0, 5, 3, 9, 6, 8, 1, 4, 2, 7], dtype=int32)
>>> hierarchy.leaves_list(hierarchy.optimal_leaf_ordering(Z, X))
array([3, 9, 0, 5, 8, 2, 7, 4, 1, 6], dtype=int32)
"""
Z = np.asarray(Z, order='c')
is_valid_linkage(Z, throw=True, name='Z')
y = _convert_to_double(np.asarray(y, order='c'))
if y.ndim == 1:
distance.is_valid_y(y, throw=True, name='y')
[y] = _copy_arrays_if_base_present([y])
elif y.ndim == 2:
if y.shape[0] == y.shape[1] and np.allclose(np.diag(y), 0):
if np.all(y >= 0) and np.allclose(y, y.T):
_warning('The symmetric non-negative hollow observation '
'matrix looks suspiciously like an uncondensed '
'distance matrix')
y = distance.pdist(y, metric)
else:
raise ValueError("`y` must be 1 or 2 dimensional.")
if not np.all(np.isfinite(y)):
raise ValueError("The condensed distance matrix must contain only "
"finite values.")
return _optimal_leaf_ordering.optimal_leaf_ordering(Z, y)
def _convert_to_bool(X):
if X.dtype != bool:
X = X.astype(bool)
if not X.flags.contiguous:
X = X.copy()
return X
def _convert_to_double(X):
if X.dtype != np.double:
X = X.astype(np.double)
if not X.flags.contiguous:
X = X.copy()
return X
def cophenet(Z, Y=None):
"""
Calculate the cophenetic distances between each observation in
the hierarchical clustering defined by the linkage ``Z``.
Suppose ``p`` and ``q`` are original observations in
disjoint clusters ``s`` and ``t``, respectively and
``s`` and ``t`` are joined by a direct parent cluster
``u``. The cophenetic distance between observations
``i`` and ``j`` is simply the distance between
clusters ``s`` and ``t``.
Parameters
----------
Z : ndarray
The hierarchical clustering encoded as an array
(see `linkage` function).
Y : ndarray (optional)
Calculates the cophenetic correlation coefficient ``c`` of a
hierarchical clustering defined by the linkage matrix `Z`
of a set of :math:`n` observations in :math:`m`
dimensions. `Y` is the condensed distance matrix from which
`Z` was generated.
Returns
-------
c : ndarray
The cophentic correlation distance (if ``Y`` is passed).
d : ndarray
The cophenetic distance matrix in condensed form. The
:math:`ij` th entry is the cophenetic distance between
original observations :math:`i` and :math:`j`.
See Also
--------
linkage: for a description of what a linkage matrix is.
scipy.spatial.distance.squareform: transforming condensed matrices into square ones.
Examples
--------
>>> from scipy.cluster.hierarchy import single, cophenet
>>> from scipy.spatial.distance import pdist, squareform
Given a dataset ``X`` and a linkage matrix ``Z``, the cophenetic distance
between two points of ``X`` is the distance between the largest two
distinct clusters that each of the points:
>>> X = [[0, 0], [0, 1], [1, 0],
... [0, 4], [0, 3], [1, 4],
... [4, 0], [3, 0], [4, 1],
... [4, 4], [3, 4], [4, 3]]
``X`` corresponds to this dataset ::
x x x x
x x
x x
x x x x
>>> Z = single(pdist(X))
>>> Z
array([[ 0., 1., 1., 2.],
[ 2., 12., 1., 3.],
[ 3., 4., 1., 2.],
[ 5., 14., 1., 3.],
[ 6., 7., 1., 2.],
[ 8., 16., 1., 3.],
[ 9., 10., 1., 2.],
[11., 18., 1., 3.],
[13., 15., 2., 6.],
[17., 20., 2., 9.],
[19., 21., 2., 12.]])
>>> cophenet(Z)
array([1., 1., 2., 2., 2., 2., 2., 2., 2., 2., 2., 1., 2., 2., 2., 2., 2.,
2., 2., 2., 2., 2., 2., 2., 2., 2., 2., 2., 2., 2., 1., 1., 2., 2.,
2., 2., 2., 2., 1., 2., 2., 2., 2., 2., 2., 2., 2., 2., 2., 2., 2.,
1., 1., 2., 2., 2., 1., 2., 2., 2., 2., 2., 2., 1., 1., 1.])
The output of the `scipy.cluster.hierarchy.cophenet` method is
represented in condensed form. We can use
`scipy.spatial.distance.squareform` to see the output as a
regular matrix (where each element ``ij`` denotes the cophenetic distance
between each ``i``, ``j`` pair of points in ``X``):
>>> squareform(cophenet(Z))
array([[0., 1., 1., 2., 2., 2., 2., 2., 2., 2., 2., 2.],
[1., 0., 1., 2., 2., 2., 2., 2., 2., 2., 2., 2.],
[1., 1., 0., 2., 2., 2., 2., 2., 2., 2., 2., 2.],
[2., 2., 2., 0., 1., 1., 2., 2., 2., 2., 2., 2.],
[2., 2., 2., 1., 0., 1., 2., 2., 2., 2., 2., 2.],
[2., 2., 2., 1., 1., 0., 2., 2., 2., 2., 2., 2.],
[2., 2., 2., 2., 2., 2., 0., 1., 1., 2., 2., 2.],
[2., 2., 2., 2., 2., 2., 1., 0., 1., 2., 2., 2.],
[2., 2., 2., 2., 2., 2., 1., 1., 0., 2., 2., 2.],
[2., 2., 2., 2., 2., 2., 2., 2., 2., 0., 1., 1.],
[2., 2., 2., 2., 2., 2., 2., 2., 2., 1., 0., 1.],
[2., 2., 2., 2., 2., 2., 2., 2., 2., 1., 1., 0.]])
In this example, the cophenetic distance between points on ``X`` that are
very close (i.e., in the same corner) is 1. For other pairs of points is 2,
because the points will be located in clusters at different
corners - thus, the distance between these clusters will be larger.
"""
Z = np.asarray(Z, order='c')
is_valid_linkage(Z, throw=True, name='Z')
Zs = Z.shape
n = Zs[0] + 1
zz = np.zeros((n * (n-1)) // 2, dtype=np.double)
# Since the C code does not support striding using strides.
# The dimensions are used instead.
Z = _convert_to_double(Z)
_hierarchy.cophenetic_distances(Z, zz, int(n))
if Y is None:
return zz
Y = np.asarray(Y, order='c')
distance.is_valid_y(Y, throw=True, name='Y')
z = zz.mean()
y = Y.mean()
Yy = Y - y
Zz = zz - z
numerator = (Yy * Zz)
denomA = Yy**2
denomB = Zz**2
c = numerator.sum() / np.sqrt((denomA.sum() * denomB.sum()))
return (c, zz)
def inconsistent(Z, d=2):
r"""
Calculate inconsistency statistics on a linkage matrix.
Parameters
----------
Z : ndarray
The :math:`(n-1)` by 4 matrix encoding the linkage (hierarchical
clustering). See `linkage` documentation for more information on its
form.
d : int, optional
The number of links up to `d` levels below each non-singleton cluster.
Returns
-------
R : ndarray
A :math:`(n-1)` by 4 matrix where the ``i``'th row contains the link
statistics for the non-singleton cluster ``i``. The link statistics are
computed over the link heights for links :math:`d` levels below the
cluster ``i``. ``R[i,0]`` and ``R[i,1]`` are the mean and standard
deviation of the link heights, respectively; ``R[i,2]`` is the number
of links included in the calculation; and ``R[i,3]`` is the
inconsistency coefficient,
.. math:: \frac{\mathtt{Z[i,2]} - \mathtt{R[i,0]}} {R[i,1]}
Notes
-----
This function behaves similarly to the MATLAB(TM) ``inconsistent``
function.
Examples
--------
>>> from scipy.cluster.hierarchy import inconsistent, linkage
>>> from matplotlib import pyplot as plt
>>> X = [[i] for i in [2, 8, 0, 4, 1, 9, 9, 0]]
>>> Z = linkage(X, 'ward')
>>> print(Z)
[[ 5. 6. 0. 2. ]
[ 2. 7. 0. 2. ]
[ 0. 4. 1. 2. ]
[ 1. 8. 1.15470054 3. ]
[ 9. 10. 2.12132034 4. ]
[ 3. 12. 4.11096096 5. ]
[11. 13. 14.07183949 8. ]]
>>> inconsistent(Z)
array([[ 0. , 0. , 1. , 0. ],
[ 0. , 0. , 1. , 0. ],
[ 1. , 0. , 1. , 0. ],
[ 0.57735027, 0.81649658, 2. , 0.70710678],
[ 1.04044011, 1.06123822, 3. , 1.01850858],
[ 3.11614065, 1.40688837, 2. , 0.70710678],
[ 6.44583366, 6.76770586, 3. , 1.12682288]])
"""
Z = np.asarray(Z, order='c')
Zs = Z.shape
is_valid_linkage(Z, throw=True, name='Z')
if (not d == np.floor(d)) or d < 0:
raise ValueError('The second argument d must be a nonnegative '
'integer value.')
# Since the C code does not support striding using strides.
# The dimensions are used instead.
[Z] = _copy_arrays_if_base_present([Z])
n = Zs[0] + 1
R = np.zeros((n - 1, 4), dtype=np.double)
_hierarchy.inconsistent(Z, R, int(n), int(d))
return R
def from_mlab_linkage(Z):
"""
Convert a linkage matrix generated by MATLAB(TM) to a new
linkage matrix compatible with this module.
The conversion does two things:
* the indices are converted from ``1..N`` to ``0..(N-1)`` form,
and
* a fourth column ``Z[:,3]`` is added where ``Z[i,3]`` represents the
number of original observations (leaves) in the non-singleton
cluster ``i``.
This function is useful when loading in linkages from legacy data
files generated by MATLAB.
Parameters
----------
Z : ndarray
A linkage matrix generated by MATLAB(TM).
Returns
-------
ZS : ndarray
A linkage matrix compatible with ``scipy.cluster.hierarchy``.
See Also
--------
linkage: for a description of what a linkage matrix is.
to_mlab_linkage: transform from SciPy to MATLAB format.
Examples
--------
>>> import numpy as np
>>> from scipy.cluster.hierarchy import ward, from_mlab_linkage
Given a linkage matrix in MATLAB format ``mZ``, we can use
`scipy.cluster.hierarchy.from_mlab_linkage` to import
it into SciPy format:
>>> mZ = np.array([[1, 2, 1], [4, 5, 1], [7, 8, 1],
... [10, 11, 1], [3, 13, 1.29099445],
... [6, 14, 1.29099445],
... [9, 15, 1.29099445],
... [12, 16, 1.29099445],
... [17, 18, 5.77350269],
... [19, 20, 5.77350269],
... [21, 22, 8.16496581]])
>>> Z = from_mlab_linkage(mZ)
>>> Z
array([[ 0. , 1. , 1. , 2. ],
[ 3. , 4. , 1. , 2. ],
[ 6. , 7. , 1. , 2. ],
[ 9. , 10. , 1. , 2. ],
[ 2. , 12. , 1.29099445, 3. ],
[ 5. , 13. , 1.29099445, 3. ],
[ 8. , 14. , 1.29099445, 3. ],
[ 11. , 15. , 1.29099445, 3. ],
[ 16. , 17. , 5.77350269, 6. ],
[ 18. , 19. , 5.77350269, 6. ],
[ 20. , 21. , 8.16496581, 12. ]])
As expected, the linkage matrix ``Z`` returned includes an
additional column counting the number of original samples in
each cluster. Also, all cluster indices are reduced by 1
(MATLAB format uses 1-indexing, whereas SciPy uses 0-indexing).
"""
Z = np.asarray(Z, dtype=np.double, order='c')
Zs = Z.shape
# If it's empty, return it.
if len(Zs) == 0 or (len(Zs) == 1 and Zs[0] == 0):
return Z.copy()
if len(Zs) != 2:
raise ValueError("The linkage array must be rectangular.")
# If it contains no rows, return it.
if Zs[0] == 0:
return Z.copy()
Zpart = Z.copy()
if Zpart[:, 0:2].min() != 1.0 and Zpart[:, 0:2].max() != 2 * Zs[0]:
raise ValueError('The format of the indices is not 1..N')
Zpart[:, 0:2] -= 1.0
CS = np.zeros((Zs[0],), dtype=np.double)
_hierarchy.calculate_cluster_sizes(Zpart, CS, int(Zs[0]) + 1)
return np.hstack([Zpart, CS.reshape(Zs[0], 1)])
def to_mlab_linkage(Z):
"""
Convert a linkage matrix to a MATLAB(TM) compatible one.
Converts a linkage matrix ``Z`` generated by the linkage function
of this module to a MATLAB(TM) compatible one. The return linkage
matrix has the last column removed and the cluster indices are
converted to ``1..N`` indexing.
Parameters
----------
Z : ndarray
A linkage matrix generated by ``scipy.cluster.hierarchy``.
Returns
-------
to_mlab_linkage : ndarray
A linkage matrix compatible with MATLAB(TM)'s hierarchical
clustering functions.
The return linkage matrix has the last column removed
and the cluster indices are converted to ``1..N`` indexing.
See Also
--------
linkage: for a description of what a linkage matrix is.
from_mlab_linkage: transform from Matlab to SciPy format.
Examples
--------
>>> from scipy.cluster.hierarchy import ward, to_mlab_linkage
>>> from scipy.spatial.distance import pdist
>>> X = [[0, 0], [0, 1], [1, 0],
... [0, 4], [0, 3], [1, 4],
... [4, 0], [3, 0], [4, 1],
... [4, 4], [3, 4], [4, 3]]
>>> Z = ward(pdist(X))
>>> Z
array([[ 0. , 1. , 1. , 2. ],
[ 3. , 4. , 1. , 2. ],
[ 6. , 7. , 1. , 2. ],
[ 9. , 10. , 1. , 2. ],
[ 2. , 12. , 1.29099445, 3. ],
[ 5. , 13. , 1.29099445, 3. ],
[ 8. , 14. , 1.29099445, 3. ],
[11. , 15. , 1.29099445, 3. ],
[16. , 17. , 5.77350269, 6. ],
[18. , 19. , 5.77350269, 6. ],
[20. , 21. , 8.16496581, 12. ]])
After a linkage matrix ``Z`` has been created, we can use
`scipy.cluster.hierarchy.to_mlab_linkage` to convert it
into MATLAB format:
>>> mZ = to_mlab_linkage(Z)
>>> mZ
array([[ 1. , 2. , 1. ],
[ 4. , 5. , 1. ],
[ 7. , 8. , 1. ],
[ 10. , 11. , 1. ],
[ 3. , 13. , 1.29099445],
[ 6. , 14. , 1.29099445],
[ 9. , 15. , 1.29099445],
[ 12. , 16. , 1.29099445],
[ 17. , 18. , 5.77350269],
[ 19. , 20. , 5.77350269],
[ 21. , 22. , 8.16496581]])
The new linkage matrix ``mZ`` uses 1-indexing for all the
clusters (instead of 0-indexing). Also, the last column of
the original linkage matrix has been dropped.
"""
Z = np.asarray(Z, order='c', dtype=np.double)
Zs = Z.shape
if len(Zs) == 0 or (len(Zs) == 1 and Zs[0] == 0):
return Z.copy()
is_valid_linkage(Z, throw=True, name='Z')
ZP = Z[:, 0:3].copy()
ZP[:, 0:2] += 1.0
return ZP
def is_monotonic(Z):
"""
Return True if the linkage passed is monotonic.
The linkage is monotonic if for every cluster :math:`s` and :math:`t`
joined, the distance between them is no less than the distance
between any previously joined clusters.
Parameters
----------
Z : ndarray
The linkage matrix to check for monotonicity.
Returns
-------
b : bool
A boolean indicating whether the linkage is monotonic.
See Also
--------
linkage: for a description of what a linkage matrix is.
Examples
--------
>>> from scipy.cluster.hierarchy import median, ward, is_monotonic
>>> from scipy.spatial.distance import pdist
By definition, some hierarchical clustering algorithms - such as
`scipy.cluster.hierarchy.ward` - produce monotonic assignments of
samples to clusters; however, this is not always true for other
hierarchical methods - e.g. `scipy.cluster.hierarchy.median`.
Given a linkage matrix ``Z`` (as the result of a hierarchical clustering
method) we can test programmatically whether it has the monotonicity
property or not, using `scipy.cluster.hierarchy.is_monotonic`:
>>> X = [[0, 0], [0, 1], [1, 0],
... [0, 4], [0, 3], [1, 4],
... [4, 0], [3, 0], [4, 1],
... [4, 4], [3, 4], [4, 3]]
>>> Z = ward(pdist(X))
>>> Z
array([[ 0. , 1. , 1. , 2. ],
[ 3. , 4. , 1. , 2. ],
[ 6. , 7. , 1. , 2. ],
[ 9. , 10. , 1. , 2. ],
[ 2. , 12. , 1.29099445, 3. ],
[ 5. , 13. , 1.29099445, 3. ],
[ 8. , 14. , 1.29099445, 3. ],
[11. , 15. , 1.29099445, 3. ],
[16. , 17. , 5.77350269, 6. ],
[18. , 19. , 5.77350269, 6. ],
[20. , 21. , 8.16496581, 12. ]])
>>> is_monotonic(Z)
True
>>> Z = median(pdist(X))
>>> Z
array([[ 0. , 1. , 1. , 2. ],
[ 3. , 4. , 1. , 2. ],
[ 9. , 10. , 1. , 2. ],
[ 6. , 7. , 1. , 2. ],
[ 2. , 12. , 1.11803399, 3. ],
[ 5. , 13. , 1.11803399, 3. ],
[ 8. , 15. , 1.11803399, 3. ],
[11. , 14. , 1.11803399, 3. ],
[18. , 19. , 3. , 6. ],
[16. , 17. , 3.5 , 6. ],
[20. , 21. , 3.25 , 12. ]])
>>> is_monotonic(Z)
False
Note that this method is equivalent to just verifying that the distances
in the third column of the linkage matrix appear in a monotonically
increasing order.
"""
Z = np.asarray(Z, order='c')
is_valid_linkage(Z, throw=True, name='Z')
# We expect the i'th value to be greater than its successor.
return (Z[1:, 2] >= Z[:-1, 2]).all()
def is_valid_im(R, warning=False, throw=False, name=None):
"""Return True if the inconsistency matrix passed is valid.
It must be a :math:`n` by 4 array of doubles. The standard
deviations ``R[:,1]`` must be nonnegative. The link counts
``R[:,2]`` must be positive and no greater than :math:`n-1`.
Parameters
----------
R : ndarray
The inconsistency matrix to check for validity.
warning : bool, optional
When True, issues a Python warning if the linkage
matrix passed is invalid.
throw : bool, optional
When True, throws a Python exception if the linkage
matrix passed is invalid.
name : str, optional
This string refers to the variable name of the invalid
linkage matrix.
Returns
-------
b : bool
True if the inconsistency matrix is valid.
See Also
--------
linkage: for a description of what a linkage matrix is.
inconsistent: for the creation of a inconsistency matrix.
Examples
--------
>>> from scipy.cluster.hierarchy import ward, inconsistent, is_valid_im
>>> from scipy.spatial.distance import pdist
Given a data set ``X``, we can apply a clustering method to obtain a
linkage matrix ``Z``. `scipy.cluster.hierarchy.inconsistent` can
be also used to obtain the inconsistency matrix ``R`` associated to
this clustering process:
>>> X = [[0, 0], [0, 1], [1, 0],
... [0, 4], [0, 3], [1, 4],
... [4, 0], [3, 0], [4, 1],
... [4, 4], [3, 4], [4, 3]]
>>> Z = ward(pdist(X))
>>> R = inconsistent(Z)
>>> Z
array([[ 0. , 1. , 1. , 2. ],
[ 3. , 4. , 1. , 2. ],
[ 6. , 7. , 1. , 2. ],
[ 9. , 10. , 1. , 2. ],
[ 2. , 12. , 1.29099445, 3. ],
[ 5. , 13. , 1.29099445, 3. ],
[ 8. , 14. , 1.29099445, 3. ],
[11. , 15. , 1.29099445, 3. ],
[16. , 17. , 5.77350269, 6. ],
[18. , 19. , 5.77350269, 6. ],
[20. , 21. , 8.16496581, 12. ]])
>>> R
array([[1. , 0. , 1. , 0. ],
[1. , 0. , 1. , 0. ],
[1. , 0. , 1. , 0. ],
[1. , 0. , 1. , 0. ],
[1.14549722, 0.20576415, 2. , 0.70710678],
[1.14549722, 0.20576415, 2. , 0.70710678],
[1.14549722, 0.20576415, 2. , 0.70710678],
[1.14549722, 0.20576415, 2. , 0.70710678],
[2.78516386, 2.58797734, 3. , 1.15470054],
[2.78516386, 2.58797734, 3. , 1.15470054],
[6.57065706, 1.38071187, 3. , 1.15470054]])
Now we can use `scipy.cluster.hierarchy.is_valid_im` to verify that
``R`` is correct:
>>> is_valid_im(R)
True
However, if ``R`` is wrongly constructed (e.g., one of the standard
deviations is set to a negative value), then the check will fail:
>>> R[-1,1] = R[-1,1] * -1
>>> is_valid_im(R)
False
"""
R = np.asarray(R, order='c')
valid = True
name_str = "%r " % name if name else ''
try:
if type(R) != np.ndarray:
raise TypeError('Variable %spassed as inconsistency matrix is not '
'a numpy array.' % name_str)
if R.dtype != np.double:
raise TypeError('Inconsistency matrix %smust contain doubles '
'(double).' % name_str)
if len(R.shape) != 2:
raise ValueError('Inconsistency matrix %smust have shape=2 (i.e. '
'be two-dimensional).' % name_str)
if R.shape[1] != 4:
raise ValueError('Inconsistency matrix %smust have 4 columns.' %
name_str)
if R.shape[0] < 1:
raise ValueError('Inconsistency matrix %smust have at least one '
'row.' % name_str)
if (R[:, 0] < 0).any():
raise ValueError('Inconsistency matrix %scontains negative link '
'height means.' % name_str)
if (R[:, 1] < 0).any():
raise ValueError('Inconsistency matrix %scontains negative link '
'height standard deviations.' % name_str)
if (R[:, 2] < 0).any():
raise ValueError('Inconsistency matrix %scontains negative link '
'counts.' % name_str)
except Exception as e:
if throw:
raise
if warning:
_warning(str(e))
valid = False
return valid
def is_valid_linkage(Z, warning=False, throw=False, name=None):
"""
Check the validity of a linkage matrix.
A linkage matrix is valid if it is a 2-D array (type double)
with :math:`n` rows and 4 columns. The first two columns must contain
indices between 0 and :math:`2n-1`. For a given row ``i``, the following
two expressions have to hold:
.. math::
0 \\leq \\mathtt{Z[i,0]} \\leq i+n-1
0 \\leq Z[i,1] \\leq i+n-1
I.e., a cluster cannot join another cluster unless the cluster being joined
has been generated.
Parameters
----------
Z : array_like
Linkage matrix.
warning : bool, optional
When True, issues a Python warning if the linkage
matrix passed is invalid.
throw : bool, optional
When True, throws a Python exception if the linkage
matrix passed is invalid.
name : str, optional
This string refers to the variable name of the invalid
linkage matrix.
Returns
-------
b : bool
True if the inconsistency matrix is valid.
See Also
--------
linkage: for a description of what a linkage matrix is.
Examples
--------
>>> from scipy.cluster.hierarchy import ward, is_valid_linkage
>>> from scipy.spatial.distance import pdist
All linkage matrices generated by the clustering methods in this module
will be valid (i.e., they will have the appropriate dimensions and the two
required expressions will hold for all the rows).
We can check this using `scipy.cluster.hierarchy.is_valid_linkage`:
>>> X = [[0, 0], [0, 1], [1, 0],
... [0, 4], [0, 3], [1, 4],
... [4, 0], [3, 0], [4, 1],
... [4, 4], [3, 4], [4, 3]]
>>> Z = ward(pdist(X))
>>> Z
array([[ 0. , 1. , 1. , 2. ],
[ 3. , 4. , 1. , 2. ],
[ 6. , 7. , 1. , 2. ],
[ 9. , 10. , 1. , 2. ],
[ 2. , 12. , 1.29099445, 3. ],
[ 5. , 13. , 1.29099445, 3. ],
[ 8. , 14. , 1.29099445, 3. ],
[11. , 15. , 1.29099445, 3. ],
[16. , 17. , 5.77350269, 6. ],
[18. , 19. , 5.77350269, 6. ],
[20. , 21. , 8.16496581, 12. ]])
>>> is_valid_linkage(Z)
True
However, if we create a linkage matrix in a wrong way - or if we modify
a valid one in a way that any of the required expressions don't hold
anymore, then the check will fail:
>>> Z[3][1] = 20 # the cluster number 20 is not defined at this point
>>> is_valid_linkage(Z)
False
"""
Z = np.asarray(Z, order='c')
valid = True
name_str = "%r " % name if name else ''
try:
if type(Z) != np.ndarray:
raise TypeError('Passed linkage argument %sis not a valid array.' %
name_str)
if Z.dtype != np.double:
raise TypeError('Linkage matrix %smust contain doubles.' % name_str)
if len(Z.shape) != 2:
raise ValueError('Linkage matrix %smust have shape=2 (i.e. be '
'two-dimensional).' % name_str)
if Z.shape[1] != 4:
raise ValueError('Linkage matrix %smust have 4 columns.' % name_str)
if Z.shape[0] == 0:
raise ValueError('Linkage must be computed on at least two '
'observations.')
n = Z.shape[0]
if n > 1:
if ((Z[:, 0] < 0).any() or (Z[:, 1] < 0).any()):
raise ValueError('Linkage %scontains negative indices.' %
name_str)
if (Z[:, 2] < 0).any():
raise ValueError('Linkage %scontains negative distances.' %
name_str)
if (Z[:, 3] < 0).any():
raise ValueError('Linkage %scontains negative counts.' %
name_str)
if _check_hierarchy_uses_cluster_before_formed(Z):
raise ValueError('Linkage %suses non-singleton cluster before '
'it is formed.' % name_str)
if _check_hierarchy_uses_cluster_more_than_once(Z):
raise ValueError('Linkage %suses the same cluster more than once.'
% name_str)
except Exception as e:
if throw:
raise
if warning:
_warning(str(e))
valid = False
return valid
def _check_hierarchy_uses_cluster_before_formed(Z):
n = Z.shape[0] + 1
for i in range(0, n - 1):
if Z[i, 0] >= n + i or Z[i, 1] >= n + i:
return True
return False
def _check_hierarchy_uses_cluster_more_than_once(Z):
n = Z.shape[0] + 1
chosen = set([])
for i in range(0, n - 1):
if (Z[i, 0] in chosen) or (Z[i, 1] in chosen) or Z[i, 0] == Z[i, 1]:
return True
chosen.add(Z[i, 0])
chosen.add(Z[i, 1])
return False
def _check_hierarchy_not_all_clusters_used(Z):
n = Z.shape[0] + 1
chosen = set([])
for i in range(0, n - 1):
chosen.add(int(Z[i, 0]))
chosen.add(int(Z[i, 1]))
must_chosen = set(range(0, 2 * n - 2))
return len(must_chosen.difference(chosen)) > 0
def num_obs_linkage(Z):
"""
Return the number of original observations of the linkage matrix passed.
Parameters
----------
Z : ndarray
The linkage matrix on which to perform the operation.
Returns
-------
n : int
The number of original observations in the linkage.
Examples
--------
>>> from scipy.cluster.hierarchy import ward, num_obs_linkage
>>> from scipy.spatial.distance import pdist
>>> X = [[0, 0], [0, 1], [1, 0],
... [0, 4], [0, 3], [1, 4],
... [4, 0], [3, 0], [4, 1],
... [4, 4], [3, 4], [4, 3]]
>>> Z = ward(pdist(X))
``Z`` is a linkage matrix obtained after using the Ward clustering method
with ``X``, a dataset with 12 data points.
>>> num_obs_linkage(Z)
12
"""
Z = np.asarray(Z, order='c')
is_valid_linkage(Z, throw=True, name='Z')
return (Z.shape[0] + 1)
def correspond(Z, Y):
"""
Check for correspondence between linkage and condensed distance matrices.
They must have the same number of original observations for
the check to succeed.
This function is useful as a sanity check in algorithms that make
extensive use of linkage and distance matrices that must
correspond to the same set of original observations.
Parameters
----------
Z : array_like
The linkage matrix to check for correspondence.
Y : array_like
The condensed distance matrix to check for correspondence.
Returns
-------
b : bool
A boolean indicating whether the linkage matrix and distance
matrix could possibly correspond to one another.
See Also
--------
linkage: for a description of what a linkage matrix is.
Examples
--------
>>> from scipy.cluster.hierarchy import ward, correspond
>>> from scipy.spatial.distance import pdist
This method can be used to check if a given linkage matrix ``Z`` has been
obtained from the application of a cluster method over a dataset ``X``:
>>> X = [[0, 0], [0, 1], [1, 0],
... [0, 4], [0, 3], [1, 4],
... [4, 0], [3, 0], [4, 1],
... [4, 4], [3, 4], [4, 3]]
>>> X_condensed = pdist(X)
>>> Z = ward(X_condensed)
Here, we can compare ``Z`` and ``X`` (in condensed form):
>>> correspond(Z, X_condensed)
True
"""
is_valid_linkage(Z, throw=True)
distance.is_valid_y(Y, throw=True)
Z = np.asarray(Z, order='c')
Y = np.asarray(Y, order='c')
return distance.num_obs_y(Y) == num_obs_linkage(Z)
def fcluster(Z, t, criterion='inconsistent', depth=2, R=None, monocrit=None):
"""
Form flat clusters from the hierarchical clustering defined by
the given linkage matrix.
Parameters
----------
Z : ndarray
The hierarchical clustering encoded with the matrix returned
by the `linkage` function.
t : scalar
For criteria 'inconsistent', 'distance' or 'monocrit',
this is the threshold to apply when forming flat clusters.
For 'maxclust' or 'maxclust_monocrit' criteria,
this would be max number of clusters requested.
criterion : str, optional
The criterion to use in forming flat clusters. This can
be any of the following values:
``inconsistent`` :
If a cluster node and all its
descendants have an inconsistent value less than or equal
to `t`, then all its leaf descendants belong to the
same flat cluster. When no non-singleton cluster meets
this criterion, every node is assigned to its own
cluster. (Default)
``distance`` :
Forms flat clusters so that the original
observations in each flat cluster have no greater a
cophenetic distance than `t`.
``maxclust`` :
Finds a minimum threshold ``r`` so that
the cophenetic distance between any two original
observations in the same flat cluster is no more than
``r`` and no more than `t` flat clusters are formed.
``monocrit`` :
Forms a flat cluster from a cluster node c
with index i when ``monocrit[j] <= t``.
For example, to threshold on the maximum mean distance
as computed in the inconsistency matrix R with a
threshold of 0.8 do::
MR = maxRstat(Z, R, 3)
fcluster(Z, t=0.8, criterion='monocrit', monocrit=MR)
``maxclust_monocrit`` :
Forms a flat cluster from a
non-singleton cluster node ``c`` when ``monocrit[i] <=
r`` for all cluster indices ``i`` below and including
``c``. ``r`` is minimized such that no more than ``t``
flat clusters are formed. monocrit must be
monotonic. For example, to minimize the threshold t on
maximum inconsistency values so that no more than 3 flat
clusters are formed, do::
MI = maxinconsts(Z, R)
fcluster(Z, t=3, criterion='maxclust_monocrit', monocrit=MI)
depth : int, optional
The maximum depth to perform the inconsistency calculation.
It has no meaning for the other criteria. Default is 2.
R : ndarray, optional
The inconsistency matrix to use for the 'inconsistent'
criterion. This matrix is computed if not provided.
monocrit : ndarray, optional
An array of length n-1. `monocrit[i]` is the
statistics upon which non-singleton i is thresholded. The
monocrit vector must be monotonic, i.e., given a node c with
index i, for all node indices j corresponding to nodes
below c, ``monocrit[i] >= monocrit[j]``.
Returns
-------
fcluster : ndarray
An array of length ``n``. ``T[i]`` is the flat cluster number to
which original observation ``i`` belongs.
See Also
--------
linkage : for information about hierarchical clustering methods work.
Examples
--------
>>> from scipy.cluster.hierarchy import ward, fcluster
>>> from scipy.spatial.distance import pdist
All cluster linkage methods - e.g., `scipy.cluster.hierarchy.ward`
generate a linkage matrix ``Z`` as their output:
>>> X = [[0, 0], [0, 1], [1, 0],
... [0, 4], [0, 3], [1, 4],
... [4, 0], [3, 0], [4, 1],
... [4, 4], [3, 4], [4, 3]]
>>> Z = ward(pdist(X))
>>> Z
array([[ 0. , 1. , 1. , 2. ],
[ 3. , 4. , 1. , 2. ],
[ 6. , 7. , 1. , 2. ],
[ 9. , 10. , 1. , 2. ],
[ 2. , 12. , 1.29099445, 3. ],
[ 5. , 13. , 1.29099445, 3. ],
[ 8. , 14. , 1.29099445, 3. ],
[11. , 15. , 1.29099445, 3. ],
[16. , 17. , 5.77350269, 6. ],
[18. , 19. , 5.77350269, 6. ],
[20. , 21. , 8.16496581, 12. ]])
This matrix represents a dendrogram, where the first and second elements
are the two clusters merged at each step, the third element is the
distance between these clusters, and the fourth element is the size of
the new cluster - the number of original data points included.
`scipy.cluster.hierarchy.fcluster` can be used to flatten the
dendrogram, obtaining as a result an assignation of the original data
points to single clusters.
This assignation mostly depends on a distance threshold ``t`` - the maximum
inter-cluster distance allowed:
>>> fcluster(Z, t=0.9, criterion='distance')
array([ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], dtype=int32)
>>> fcluster(Z, t=1.1, criterion='distance')
array([1, 1, 2, 3, 3, 4, 5, 5, 6, 7, 7, 8], dtype=int32)
>>> fcluster(Z, t=3, criterion='distance')
array([1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4], dtype=int32)
>>> fcluster(Z, t=9, criterion='distance')
array([1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], dtype=int32)
In the first case, the threshold ``t`` is too small to allow any two
samples in the data to form a cluster, so 12 different clusters are
returned.
In the second case, the threshold is large enough to allow the first
4 points to be merged with their nearest neighbors. So, here, only 8
clusters are returned.
The third case, with a much higher threshold, allows for up to 8 data
points to be connected - so 4 clusters are returned here.
Lastly, the threshold of the fourth case is large enough to allow for
all data points to be merged together - so a single cluster is returned.
"""
Z = np.asarray(Z, order='c')
is_valid_linkage(Z, throw=True, name='Z')
n = Z.shape[0] + 1
T = np.zeros((n,), dtype='i')
# Since the C code does not support striding using strides.
# The dimensions are used instead.
[Z] = _copy_arrays_if_base_present([Z])
if criterion == 'inconsistent':
if R is None:
R = inconsistent(Z, depth)
else:
R = np.asarray(R, order='c')
is_valid_im(R, throw=True, name='R')
# Since the C code does not support striding using strides.
# The dimensions are used instead.
[R] = _copy_arrays_if_base_present([R])
_hierarchy.cluster_in(Z, R, T, float(t), int(n))
elif criterion == 'distance':
_hierarchy.cluster_dist(Z, T, float(t), int(n))
elif criterion == 'maxclust':
_hierarchy.cluster_maxclust_dist(Z, T, int(n), int(t))
elif criterion == 'monocrit':
[monocrit] = _copy_arrays_if_base_present([monocrit])
_hierarchy.cluster_monocrit(Z, monocrit, T, float(t), int(n))
elif criterion == 'maxclust_monocrit':
[monocrit] = _copy_arrays_if_base_present([monocrit])
_hierarchy.cluster_maxclust_monocrit(Z, monocrit, T, int(n), int(t))
else:
raise ValueError('Invalid cluster formation criterion: %s'
% str(criterion))
return T
def fclusterdata(X, t, criterion='inconsistent',
metric='euclidean', depth=2, method='single', R=None):
"""
Cluster observation data using a given metric.
Clusters the original observations in the n-by-m data
matrix X (n observations in m dimensions), using the euclidean
distance metric to calculate distances between original observations,
performs hierarchical clustering using the single linkage algorithm,
and forms flat clusters using the inconsistency method with `t` as the
cut-off threshold.
A 1-D array ``T`` of length ``n`` is returned. ``T[i]`` is
the index of the flat cluster to which the original observation ``i``
belongs.
Parameters
----------
X : (N, M) ndarray
N by M data matrix with N observations in M dimensions.
t : scalar
For criteria 'inconsistent', 'distance' or 'monocrit',
this is the threshold to apply when forming flat clusters.
For 'maxclust' or 'maxclust_monocrit' criteria,
this would be max number of clusters requested.
criterion : str, optional
Specifies the criterion for forming flat clusters. Valid
values are 'inconsistent' (default), 'distance', or 'maxclust'
cluster formation algorithms. See `fcluster` for descriptions.
metric : str or function, optional
The distance metric for calculating pairwise distances. See
``distance.pdist`` for descriptions and linkage to verify
compatibility with the linkage method.
depth : int, optional
The maximum depth for the inconsistency calculation. See
`inconsistent` for more information.
method : str, optional
The linkage method to use (single, complete, average,
weighted, median centroid, ward). See `linkage` for more
information. Default is "single".
R : ndarray, optional
The inconsistency matrix. It will be computed if necessary
if it is not passed.
Returns
-------
fclusterdata : ndarray
A vector of length n. T[i] is the flat cluster number to
which original observation i belongs.
See Also
--------
scipy.spatial.distance.pdist : pairwise distance metrics
Notes
-----
This function is similar to the MATLAB function ``clusterdata``.
Examples
--------
>>> from scipy.cluster.hierarchy import fclusterdata
This is a convenience method that abstracts all the steps to perform in a
typical SciPy's hierarchical clustering workflow.
* Transform the input data into a condensed matrix with `scipy.spatial.distance.pdist`.
* Apply a clustering method.
* Obtain flat clusters at a user defined distance threshold ``t`` using `scipy.cluster.hierarchy.fcluster`.
>>> X = [[0, 0], [0, 1], [1, 0],
... [0, 4], [0, 3], [1, 4],
... [4, 0], [3, 0], [4, 1],
... [4, 4], [3, 4], [4, 3]]
>>> fclusterdata(X, t=1)
array([3, 3, 3, 4, 4, 4, 2, 2, 2, 1, 1, 1], dtype=int32)
The output here (for the dataset ``X``, distance threshold ``t``, and the
default settings) is four clusters with three data points each.
"""
X = np.asarray(X, order='c', dtype=np.double)
if type(X) != np.ndarray or len(X.shape) != 2:
raise TypeError('The observation matrix X must be an n by m numpy '
'array.')
Y = distance.pdist(X, metric=metric)
Z = linkage(Y, method=method)
if R is None:
R = inconsistent(Z, d=depth)
else:
R = np.asarray(R, order='c')
T = fcluster(Z, criterion=criterion, depth=depth, R=R, t=t)
return T
def leaves_list(Z):
"""
Return a list of leaf node ids.
The return corresponds to the observation vector index as it appears
in the tree from left to right. Z is a linkage matrix.
Parameters
----------
Z : ndarray
The hierarchical clustering encoded as a matrix. `Z` is
a linkage matrix. See `linkage` for more information.
Returns
-------
leaves_list : ndarray
The list of leaf node ids.
See Also
--------
dendrogram: for information about dendrogram structure.
Examples
--------
>>> from scipy.cluster.hierarchy import ward, dendrogram, leaves_list
>>> from scipy.spatial.distance import pdist
>>> from matplotlib import pyplot as plt
>>> X = [[0, 0], [0, 1], [1, 0],
... [0, 4], [0, 3], [1, 4],
... [4, 0], [3, 0], [4, 1],
... [4, 4], [3, 4], [4, 3]]
>>> Z = ward(pdist(X))
The linkage matrix ``Z`` represents a dendrogram, that is, a tree that
encodes the structure of the clustering performed.
`scipy.cluster.hierarchy.leaves_list` shows the mapping between
indices in the ``X`` dataset and leaves in the dendrogram:
>>> leaves_list(Z)
array([ 2, 0, 1, 5, 3, 4, 8, 6, 7, 11, 9, 10], dtype=int32)
>>> fig = plt.figure(figsize=(25, 10))
>>> dn = dendrogram(Z)
>>> plt.show()
"""
Z = np.asarray(Z, order='c')
is_valid_linkage(Z, throw=True, name='Z')
n = Z.shape[0] + 1
ML = np.zeros((n,), dtype='i')
[Z] = _copy_arrays_if_base_present([Z])
_hierarchy.prelist(Z, ML, int(n))
return ML
# Maps number of leaves to text size.
#
# p <= 20, size="12"
# 20 < p <= 30, size="10"
# 30 < p <= 50, size="8"
# 50 < p <= np.inf, size="6"
_dtextsizes = {20: 12, 30: 10, 50: 8, 85: 6, np.inf: 5}
_drotation = {20: 0, 40: 45, np.inf: 90}
_dtextsortedkeys = list(_dtextsizes.keys())
_dtextsortedkeys.sort()
_drotationsortedkeys = list(_drotation.keys())
_drotationsortedkeys.sort()
def _remove_dups(L):
"""
Remove duplicates AND preserve the original order of the elements.
The set class is not guaranteed to do this.
"""
seen_before = set([])
L2 = []
for i in L:
if i not in seen_before:
seen_before.add(i)
L2.append(i)
return L2
def _get_tick_text_size(p):
for k in _dtextsortedkeys:
if p <= k:
return _dtextsizes[k]
def _get_tick_rotation(p):
for k in _drotationsortedkeys:
if p <= k:
return _drotation[k]
def _plot_dendrogram(icoords, dcoords, ivl, p, n, mh, orientation,
no_labels, color_list, leaf_font_size=None,
leaf_rotation=None, contraction_marks=None,
ax=None, above_threshold_color='C0'):
# Import matplotlib here so that it's not imported unless dendrograms
# are plotted. Raise an informative error if importing fails.
try:
# if an axis is provided, don't use pylab at all
if ax is None:
import matplotlib.pylab
import matplotlib.patches
import matplotlib.collections
except ImportError as e:
raise ImportError("You must install the matplotlib library to plot "
"the dendrogram. Use no_plot=True to calculate the "
"dendrogram without plotting.") from e
if ax is None:
ax = matplotlib.pylab.gca()
# if we're using pylab, we want to trigger a draw at the end
trigger_redraw = True
else:
trigger_redraw = False
# Independent variable plot width
ivw = len(ivl) * 10
# Dependent variable plot height
dvw = mh + mh * 0.05
iv_ticks = np.arange(5, len(ivl) * 10 + 5, 10)
if orientation in ('top', 'bottom'):
if orientation == 'top':
ax.set_ylim([0, dvw])
ax.set_xlim([0, ivw])
else:
ax.set_ylim([dvw, 0])
ax.set_xlim([0, ivw])
xlines = icoords
ylines = dcoords
if no_labels:
ax.set_xticks([])
ax.set_xticklabels([])
else:
ax.set_xticks(iv_ticks)
if orientation == 'top':
ax.xaxis.set_ticks_position('bottom')
else:
ax.xaxis.set_ticks_position('top')
# Make the tick marks invisible because they cover up the links
for line in ax.get_xticklines():
line.set_visible(False)
leaf_rot = (float(_get_tick_rotation(len(ivl)))
if (leaf_rotation is None) else leaf_rotation)
leaf_font = (float(_get_tick_text_size(len(ivl)))
if (leaf_font_size is None) else leaf_font_size)
ax.set_xticklabels(ivl, rotation=leaf_rot, size=leaf_font)
elif orientation in ('left', 'right'):
if orientation == 'left':
ax.set_xlim([dvw, 0])
ax.set_ylim([0, ivw])
else:
ax.set_xlim([0, dvw])
ax.set_ylim([0, ivw])
xlines = dcoords
ylines = icoords
if no_labels:
ax.set_yticks([])
ax.set_yticklabels([])
else:
ax.set_yticks(iv_ticks)
if orientation == 'left':
ax.yaxis.set_ticks_position('right')
else:
ax.yaxis.set_ticks_position('left')
# Make the tick marks invisible because they cover up the links
for line in ax.get_yticklines():
line.set_visible(False)
leaf_font = (float(_get_tick_text_size(len(ivl)))
if (leaf_font_size is None) else leaf_font_size)
if leaf_rotation is not None:
ax.set_yticklabels(ivl, rotation=leaf_rotation, size=leaf_font)
else:
ax.set_yticklabels(ivl, size=leaf_font)
# Let's use collections instead. This way there is a separate legend item
# for each tree grouping, rather than stupidly one for each line segment.
colors_used = _remove_dups(color_list)
color_to_lines = {}
for color in colors_used:
color_to_lines[color] = []
for (xline, yline, color) in zip(xlines, ylines, color_list):
color_to_lines[color].append(list(zip(xline, yline)))
colors_to_collections = {}
# Construct the collections.
for color in colors_used:
coll = matplotlib.collections.LineCollection(color_to_lines[color],
colors=(color,))
colors_to_collections[color] = coll
# Add all the groupings below the color threshold.
for color in colors_used:
if color != above_threshold_color:
ax.add_collection(colors_to_collections[color])
# If there's a grouping of links above the color threshold, it goes last.
if above_threshold_color in colors_to_collections:
ax.add_collection(colors_to_collections[above_threshold_color])
if contraction_marks is not None:
Ellipse = matplotlib.patches.Ellipse
for (x, y) in contraction_marks:
if orientation in ('left', 'right'):
e = Ellipse((y, x), width=dvw / 100, height=1.0)
else:
e = Ellipse((x, y), width=1.0, height=dvw / 100)
ax.add_artist(e)
e.set_clip_box(ax.bbox)
e.set_alpha(0.5)
e.set_facecolor('k')
if trigger_redraw:
matplotlib.pylab.draw_if_interactive()
# C0 is used for above threshhold color
_link_line_colors_default = ('C1', 'C2', 'C3', 'C4', 'C5', 'C6', 'C7', 'C8', 'C9')
_link_line_colors = list(_link_line_colors_default)
def set_link_color_palette(palette):
"""
Set list of matplotlib color codes for use by dendrogram.
Note that this palette is global (i.e., setting it once changes the colors
for all subsequent calls to `dendrogram`) and that it affects only the
the colors below ``color_threshold``.
Note that `dendrogram` also accepts a custom coloring function through its
``link_color_func`` keyword, which is more flexible and non-global.
Parameters
----------
palette : list of str or None
A list of matplotlib color codes. The order of the color codes is the
order in which the colors are cycled through when color thresholding in
the dendrogram.
If ``None``, resets the palette to its default (which are matplotlib
default colors C1 to C9).
Returns
-------
None
See Also
--------
dendrogram
Notes
-----
Ability to reset the palette with ``None`` added in SciPy 0.17.0.
Examples
--------
>>> from scipy.cluster import hierarchy
>>> ytdist = np.array([662., 877., 255., 412., 996., 295., 468., 268.,
... 400., 754., 564., 138., 219., 869., 669.])
>>> Z = hierarchy.linkage(ytdist, 'single')
>>> dn = hierarchy.dendrogram(Z, no_plot=True)
>>> dn['color_list']
['C1', 'C0', 'C0', 'C0', 'C0']
>>> hierarchy.set_link_color_palette(['c', 'm', 'y', 'k'])
>>> dn = hierarchy.dendrogram(Z, no_plot=True, above_threshold_color='b')
>>> dn['color_list']
['c', 'b', 'b', 'b', 'b']
>>> dn = hierarchy.dendrogram(Z, no_plot=True, color_threshold=267,
... above_threshold_color='k')
>>> dn['color_list']
['c', 'm', 'm', 'k', 'k']
Now, reset the color palette to its default:
>>> hierarchy.set_link_color_palette(None)
"""
if palette is None:
# reset to its default
palette = _link_line_colors_default
elif type(palette) not in (list, tuple):
raise TypeError("palette must be a list or tuple")
_ptypes = [isinstance(p, str) for p in palette]
if False in _ptypes:
raise TypeError("all palette list elements must be color strings")
global _link_line_colors
_link_line_colors = palette
def dendrogram(Z, p=30, truncate_mode=None, color_threshold=None,
get_leaves=True, orientation='top', labels=None,
count_sort=False, distance_sort=False, show_leaf_counts=True,
no_plot=False, no_labels=False, leaf_font_size=None,
leaf_rotation=None, leaf_label_func=None,
show_contracted=False, link_color_func=None, ax=None,
above_threshold_color='C0'):
"""
Plot the hierarchical clustering as a dendrogram.
The dendrogram illustrates how each cluster is
composed by drawing a U-shaped link between a non-singleton
cluster and its children. The top of the U-link indicates a
cluster merge. The two legs of the U-link indicate which clusters
were merged. The length of the two legs of the U-link represents
the distance between the child clusters. It is also the
cophenetic distance between original observations in the two
children clusters. Customized to track cluster index traversal in
construction of dendrogram for labeling purposes.
Parameters
----------
Z : ndarray
The linkage matrix encoding the hierarchical clustering to
render as a dendrogram. See the ``linkage`` function for more
information on the format of ``Z``.
p : int, optional
The ``p`` parameter for ``truncate_mode``.
truncate_mode : str, optional
The dendrogram can be hard to read when the original
observation matrix from which the linkage is derived is
large. Truncation is used to condense the dendrogram. There
are several modes:
``None``
No truncation is performed (default).
Note: ``'none'`` is an alias for ``None`` that's kept for
backward compatibility.
``'lastp'``
The last ``p`` non-singleton clusters formed in the linkage are the
only non-leaf nodes in the linkage; they correspond to rows
``Z[n-p-2:end]`` in ``Z``. All other non-singleton clusters are
contracted into leaf nodes.
``'level'``
No more than ``p`` levels of the dendrogram tree are displayed.
A "level" includes all nodes with ``p`` merges from the last merge.
Note: ``'mtica'`` is an alias for ``'level'`` that's kept for
backward compatibility.
color_threshold : double, optional
For brevity, let :math:`t` be the ``color_threshold``.
Colors all the descendent links below a cluster node
:math:`k` the same color if :math:`k` is the first node below
the cut threshold :math:`t`. All links connecting nodes with
distances greater than or equal to the threshold are colored
with de default matplotlib color ``'C0'``. If :math:`t` is less
than or equal to zero, all nodes are colored ``'C0'``.
If ``color_threshold`` is None or 'default',
corresponding with MATLAB(TM) behavior, the threshold is set to
``0.7*max(Z[:,2])``.
get_leaves : bool, optional
Includes a list ``R['leaves']=H`` in the result
dictionary. For each :math:`i`, ``H[i] == j``, cluster node
``j`` appears in position ``i`` in the left-to-right traversal
of the leaves, where :math:`j < 2n-1` and :math:`i < n`.
orientation : str, optional
The direction to plot the dendrogram, which can be any
of the following strings:
``'top'``
Plots the root at the top, and plot descendent links going downwards.
(default).
``'bottom'``
Plots the root at the bottom, and plot descendent links going
upwards.
``'left'``
Plots the root at the left, and plot descendent links going right.
``'right'``
Plots the root at the right, and plot descendent links going left.
labels : ndarray, optional
By default, ``labels`` is None so the index of the original observation
is used to label the leaf nodes. Otherwise, this is an :math:`n`-sized
sequence, with ``n == Z.shape[0] + 1``. The ``labels[i]`` value is the
text to put under the :math:`i` th leaf node only if it corresponds to
an original observation and not a non-singleton cluster.
count_sort : str or bool, optional
For each node n, the order (visually, from left-to-right) n's
two descendent links are plotted is determined by this
parameter, which can be any of the following values:
``False``
Nothing is done.
``'ascending'`` or ``True``
The child with the minimum number of original objects in its cluster
is plotted first.
``'descending'``
The child with the maximum number of original objects in its cluster
is plotted first.
Note, ``distance_sort`` and ``count_sort`` cannot both be True.
distance_sort : str or bool, optional
For each node n, the order (visually, from left-to-right) n's
two descendent links are plotted is determined by this
parameter, which can be any of the following values:
``False``
Nothing is done.
``'ascending'`` or ``True``
The child with the minimum distance between its direct descendents is
plotted first.
``'descending'``
The child with the maximum distance between its direct descendents is
plotted first.
Note ``distance_sort`` and ``count_sort`` cannot both be True.
show_leaf_counts : bool, optional
When True, leaf nodes representing :math:`k>1` original
observation are labeled with the number of observations they
contain in parentheses.
no_plot : bool, optional
When True, the final rendering is not performed. This is
useful if only the data structures computed for the rendering
are needed or if matplotlib is not available.
no_labels : bool, optional
When True, no labels appear next to the leaf nodes in the
rendering of the dendrogram.
leaf_rotation : double, optional
Specifies the angle (in degrees) to rotate the leaf
labels. When unspecified, the rotation is based on the number of
nodes in the dendrogram (default is 0).
leaf_font_size : int, optional
Specifies the font size (in points) of the leaf labels. When
unspecified, the size based on the number of nodes in the
dendrogram.
leaf_label_func : lambda or function, optional
When leaf_label_func is a callable function, for each
leaf with cluster index :math:`k < 2n-1`. The function
is expected to return a string with the label for the
leaf.
Indices :math:`k < n` correspond to original observations
while indices :math:`k \\geq n` correspond to non-singleton
clusters.
For example, to label singletons with their node id and
non-singletons with their id, count, and inconsistency
coefficient, simply do::
# First define the leaf label function.
def llf(id):
if id < n:
return str(id)
else:
return '[%d %d %1.2f]' % (id, count, R[n-id,3])
# The text for the leaf nodes is going to be big so force
# a rotation of 90 degrees.
dendrogram(Z, leaf_label_func=llf, leaf_rotation=90)
show_contracted : bool, optional
When True the heights of non-singleton nodes contracted
into a leaf node are plotted as crosses along the link
connecting that leaf node. This really is only useful when
truncation is used (see ``truncate_mode`` parameter).
link_color_func : callable, optional
If given, `link_color_function` is called with each non-singleton id
corresponding to each U-shaped link it will paint. The function is
expected to return the color to paint the link, encoded as a matplotlib
color string code. For example::
dendrogram(Z, link_color_func=lambda k: colors[k])
colors the direct links below each untruncated non-singleton node
``k`` using ``colors[k]``.
ax : matplotlib Axes instance, optional
If None and `no_plot` is not True, the dendrogram will be plotted
on the current axes. Otherwise if `no_plot` is not True the
dendrogram will be plotted on the given ``Axes`` instance. This can be
useful if the dendrogram is part of a more complex figure.
above_threshold_color : str, optional
This matplotlib color string sets the color of the links above the
color_threshold. The default is ``'C0'``.
Returns
-------
R : dict
A dictionary of data structures computed to render the
dendrogram. Its has the following keys:
``'color_list'``
A list of color names. The k'th element represents the color of the
k'th link.
``'icoord'`` and ``'dcoord'``
Each of them is a list of lists. Let ``icoord = [I1, I2, ..., Ip]``
where ``Ik = [xk1, xk2, xk3, xk4]`` and ``dcoord = [D1, D2, ..., Dp]``
where ``Dk = [yk1, yk2, yk3, yk4]``, then the k'th link painted is
``(xk1, yk1)`` - ``(xk2, yk2)`` - ``(xk3, yk3)`` - ``(xk4, yk4)``.
``'ivl'``
A list of labels corresponding to the leaf nodes.
``'leaves'``
For each i, ``H[i] == j``, cluster node ``j`` appears in position
``i`` in the left-to-right traversal of the leaves, where
:math:`j < 2n-1` and :math:`i < n`. If ``j`` is less than ``n``, the
``i``-th leaf node corresponds to an original observation.
Otherwise, it corresponds to a non-singleton cluster.
``'leaves_color_list'``
A list of color names. The k'th element represents the color of the
k'th leaf.
See Also
--------
linkage, set_link_color_palette
Notes
-----
It is expected that the distances in ``Z[:,2]`` be monotonic, otherwise
crossings appear in the dendrogram.
Examples
--------
>>> from scipy.cluster import hierarchy
>>> import matplotlib.pyplot as plt
A very basic example:
>>> ytdist = np.array([662., 877., 255., 412., 996., 295., 468., 268.,
... 400., 754., 564., 138., 219., 869., 669.])
>>> Z = hierarchy.linkage(ytdist, 'single')
>>> plt.figure()
>>> dn = hierarchy.dendrogram(Z)
Now, plot in given axes, improve the color scheme and use both vertical and
horizontal orientations:
>>> hierarchy.set_link_color_palette(['m', 'c', 'y', 'k'])
>>> fig, axes = plt.subplots(1, 2, figsize=(8, 3))
>>> dn1 = hierarchy.dendrogram(Z, ax=axes[0], above_threshold_color='y',
... orientation='top')
>>> dn2 = hierarchy.dendrogram(Z, ax=axes[1],
... above_threshold_color='#bcbddc',
... orientation='right')
>>> hierarchy.set_link_color_palette(None) # reset to default after use
>>> plt.show()
"""
# This feature was thought about but never implemented (still useful?):
#
# ... = dendrogram(..., leaves_order=None)
#
# Plots the leaves in the order specified by a vector of
# original observation indices. If the vector contains duplicates
# or results in a crossing, an exception will be thrown. Passing
# None orders leaf nodes based on the order they appear in the
# pre-order traversal.
Z = np.asarray(Z, order='c')
if orientation not in ["top", "left", "bottom", "right"]:
raise ValueError("orientation must be one of 'top', 'left', "
"'bottom', or 'right'")
if labels is not None and Z.shape[0] + 1 != len(labels):
raise ValueError("Dimensions of Z and labels must be consistent.")
is_valid_linkage(Z, throw=True, name='Z')
Zs = Z.shape
n = Zs[0] + 1
if type(p) in (int, float):
p = int(p)
else:
raise TypeError('The second argument must be a number')
if truncate_mode not in ('lastp', 'mlab', 'mtica', 'level', 'none', None):
# 'mlab' and 'mtica' are kept working for backwards compat.
raise ValueError('Invalid truncation mode.')
if truncate_mode == 'lastp' or truncate_mode == 'mlab':
if p > n or p == 0:
p = n
if truncate_mode == 'mtica':
# 'mtica' is an alias
truncate_mode = 'level'
if truncate_mode == 'level':
if p <= 0:
p = np.inf
if get_leaves:
lvs = []
else:
lvs = None
icoord_list = []
dcoord_list = []
color_list = []
current_color = [0]
currently_below_threshold = [False]
ivl = [] # list of leaves
if color_threshold is None or (isinstance(color_threshold, str) and
color_threshold == 'default'):
color_threshold = max(Z[:, 2]) * 0.7
R = {'icoord': icoord_list, 'dcoord': dcoord_list, 'ivl': ivl,
'leaves': lvs, 'color_list': color_list}
# Empty list will be filled in _dendrogram_calculate_info
contraction_marks = [] if show_contracted else None
traversal = []
_dendrogram_calculate_info(
Z=Z, p=p,
truncate_mode=truncate_mode,
color_threshold=color_threshold,
get_leaves=get_leaves,
orientation=orientation,
labels=labels,
count_sort=count_sort,
distance_sort=distance_sort,
show_leaf_counts=show_leaf_counts,
i=2*n - 2,
iv=0.0,
ivl=ivl,
n=n,
icoord_list=icoord_list,
dcoord_list=dcoord_list,
lvs=lvs,
current_color=current_color,
color_list=color_list,
currently_below_threshold=currently_below_threshold,
leaf_label_func=leaf_label_func,
contraction_marks=contraction_marks,
link_color_func=link_color_func,
above_threshold_color=above_threshold_color,
traversal=traversal)
if not no_plot:
mh = max(Z[:, 2])
_plot_dendrogram(icoord_list, dcoord_list, ivl, p, n, mh, orientation,
no_labels, color_list,
leaf_font_size=leaf_font_size,
leaf_rotation=leaf_rotation,
contraction_marks=contraction_marks,
ax=ax,
above_threshold_color=above_threshold_color)
R["leaves_color_list"] = _get_leaves_color_list(R)
R['traversal'] = traversal
return R
def _get_leaves_color_list(R):
leaves_color_list = [None] * len(R['leaves'])
for link_x, link_y, link_color in zip(R['icoord'],
R['dcoord'],
R['color_list']):
for (xi, yi) in zip(link_x, link_y):
if yi == 0.0: # if yi is 0.0, the point is a leaf
# xi of leaves are 5, 15, 25, 35, ... (see `iv_ticks`)
# index of leaves are 0, 1, 2, 3, ... as below
leaf_index = (int(xi) - 5) // 10
# each leaf has a same color of its link.
leaves_color_list[leaf_index] = link_color
return leaves_color_list
def _append_singleton_leaf_node(Z, p, n, level, lvs, ivl, leaf_label_func,
i, labels):
# If the leaf id structure is not None and is a list then the caller
# to dendrogram has indicated that cluster id's corresponding to the
# leaf nodes should be recorded.
if lvs is not None:
lvs.append(int(i))
# If leaf node labels are to be displayed...
if ivl is not None:
# If a leaf_label_func has been provided, the label comes from the
# string returned from the leaf_label_func, which is a function
# passed to dendrogram.
if leaf_label_func:
ivl.append(leaf_label_func(int(i)))
else:
# Otherwise, if the dendrogram caller has passed a labels list
# for the leaf nodes, use it.
if labels is not None:
ivl.append(labels[int(i - n)])
else:
# Otherwise, use the id as the label for the leaf.x
ivl.append(str(int(i)))
def _append_nonsingleton_leaf_node(Z, p, n, level, lvs, ivl, leaf_label_func,
i, labels, show_leaf_counts):
# If the leaf id structure is not None and is a list then the caller
# to dendrogram has indicated that cluster id's corresponding to the
# leaf nodes should be recorded.
if lvs is not None:
lvs.append(int(i))
if ivl is not None:
if leaf_label_func:
ivl.append(leaf_label_func(int(i)))
else:
if show_leaf_counts:
ivl.append("(" + str(int(Z[i - n, 3])) + ")")
else:
ivl.append("")
def _append_contraction_marks(Z, iv, i, n, contraction_marks):
_append_contraction_marks_sub(Z, iv, int(Z[i - n, 0]), n, contraction_marks)
_append_contraction_marks_sub(Z, iv, int(Z[i - n, 1]), n, contraction_marks)
def _append_contraction_marks_sub(Z, iv, i, n, contraction_marks):
if i >= n:
contraction_marks.append((iv, Z[i - n, 2]))
_append_contraction_marks_sub(Z, iv, int(Z[i - n, 0]), n, contraction_marks)
_append_contraction_marks_sub(Z, iv, int(Z[i - n, 1]), n, contraction_marks)
def _dendrogram_calculate_info(Z, p, truncate_mode,
color_threshold=np.inf, get_leaves=True,
orientation='top', labels=None,
count_sort=False, distance_sort=False,
show_leaf_counts=False, i=-1, iv=0.0,
ivl=[], n=0, icoord_list=[], dcoord_list=[],
lvs=None, mhr=False,
current_color=[], color_list=[],
currently_below_threshold=[],
leaf_label_func=None, level=0,
contraction_marks=None,
link_color_func=None,
above_threshold_color='C0',
traversal=[]):
"""
Calculate the endpoints of the links as well as the labels for the
the dendrogram rooted at the node with index i. iv is the independent
variable value to plot the left-most leaf node below the root node i
(if orientation='top', this would be the left-most x value where the
plotting of this root node i and its descendents should begin).
ivl is a list to store the labels of the leaf nodes. The leaf_label_func
is called whenever ivl != None, labels == None, and
leaf_label_func != None. When ivl != None and labels != None, the
labels list is used only for labeling the leaf nodes. When
ivl == None, no labels are generated for leaf nodes.
When get_leaves==True, a list of leaves is built as they are visited
in the dendrogram.
Returns a tuple with l being the independent variable coordinate that
corresponds to the midpoint of cluster to the left of cluster i if
i is non-singleton, otherwise the independent coordinate of the leaf
node if i is a leaf node.
Returns
-------
A tuple (left, w, h, md), where:
* left is the independent variable coordinate of the center of the
the U of the subtree
* w is the amount of space used for the subtree (in independent
variable units)
* h is the height of the subtree in dependent variable units
* md is the ``max(Z[*,2]``) for all nodes ``*`` below and including
the target node.
"""
traversal.append(i)
if n == 0:
raise ValueError("Invalid singleton cluster count n.")
if i == -1:
raise ValueError("Invalid root cluster index i.")
if truncate_mode == 'lastp':
# If the node is a leaf node but corresponds to a non-singleton
# cluster, its label is either the empty string or the number of
# original observations belonging to cluster i.
if 2*n - p > i >= n:
d = Z[i - n, 2]
_append_nonsingleton_leaf_node(Z, p, n, level, lvs, ivl,
leaf_label_func, i, labels,
show_leaf_counts)
if contraction_marks is not None:
_append_contraction_marks(Z, iv + 5.0, i, n, contraction_marks)
return (iv + 5.0, 10.0, 0.0, d)
elif i < n:
_append_singleton_leaf_node(Z, p, n, level, lvs, ivl,
leaf_label_func, i, labels)
return (iv + 5.0, 10.0, 0.0, 0.0)
elif truncate_mode == 'level':
if i > n and level > p:
d = Z[i - n, 2]
_append_nonsingleton_leaf_node(Z, p, n, level, lvs, ivl,
leaf_label_func, i, labels,
show_leaf_counts)
if contraction_marks is not None:
_append_contraction_marks(Z, iv + 5.0, i, n, contraction_marks)
return (iv + 5.0, 10.0, 0.0, d)
elif i < n:
_append_singleton_leaf_node(Z, p, n, level, lvs, ivl,
leaf_label_func, i, labels)
return (iv + 5.0, 10.0, 0.0, 0.0)
elif truncate_mode in ('mlab',):
msg = "Mode 'mlab' is deprecated in scipy 0.19.0 (it never worked)."
warnings.warn(msg, DeprecationWarning)
# Otherwise, only truncate if we have a leaf node.
#
# Only place leaves if they correspond to original observations.
if i < n:
_append_singleton_leaf_node(Z, p, n, level, lvs, ivl,
leaf_label_func, i, labels)
return (iv + 5.0, 10.0, 0.0, 0.0)
# !!! Otherwise, we don't have a leaf node, so work on plotting a
# non-leaf node.
# Actual indices of a and b
aa = int(Z[i - n, 0])
ab = int(Z[i - n, 1])
if aa >= n:
# The number of singletons below cluster a
na = Z[aa - n, 3]
# The distance between a's two direct children.
da = Z[aa - n, 2]
else:
na = 1
da = 0.0
if ab >= n:
nb = Z[ab - n, 3]
db = Z[ab - n, 2]
else:
nb = 1
db = 0.0
if count_sort == 'ascending' or count_sort == True:
# If a has a count greater than b, it and its descendents should
# be drawn to the right. Otherwise, to the left.
if na > nb:
# The cluster index to draw to the left (ua) will be ab
# and the one to draw to the right (ub) will be aa
ua = ab
ub = aa
else:
ua = aa
ub = ab
elif count_sort == 'descending':
# If a has a count less than or equal to b, it and its
# descendents should be drawn to the left. Otherwise, to
# the right.
if na > nb:
ua = aa
ub = ab
else:
ua = ab
ub = aa
elif distance_sort == 'ascending' or distance_sort == True:
# If a has a distance greater than b, it and its descendents should
# be drawn to the right. Otherwise, to the left.
if da > db:
ua = ab
ub = aa
else:
ua = aa
ub = ab
elif distance_sort == 'descending':
# If a has a distance less than or equal to b, it and its
# descendents should be drawn to the left. Otherwise, to
# the right.
if da > db:
ua = aa
ub = ab
else:
ua = ab
ub = aa
else:
ua = aa
ub = ab
# Updated iv variable and the amount of space used.
(uiva, uwa, uah, uamd) = \
_dendrogram_calculate_info(
Z=Z, p=p,
truncate_mode=truncate_mode,
color_threshold=color_threshold,
get_leaves=get_leaves,
orientation=orientation,
labels=labels,
count_sort=count_sort,
distance_sort=distance_sort,
show_leaf_counts=show_leaf_counts,
i=ua, iv=iv, ivl=ivl, n=n,
icoord_list=icoord_list,
dcoord_list=dcoord_list, lvs=lvs,
current_color=current_color,
color_list=color_list,
currently_below_threshold=currently_below_threshold,
leaf_label_func=leaf_label_func,
level=level + 1, contraction_marks=contraction_marks,
link_color_func=link_color_func,
above_threshold_color=above_threshold_color,
traversal=traversal)
h = Z[i - n, 2]
if h >= color_threshold or color_threshold <= 0:
c = above_threshold_color
if currently_below_threshold[0]:
current_color[0] = (current_color[0] + 1) % len(_link_line_colors)
currently_below_threshold[0] = False
else:
currently_below_threshold[0] = True
c = _link_line_colors[current_color[0]]
(uivb, uwb, ubh, ubmd) = \
_dendrogram_calculate_info(
Z=Z, p=p,
truncate_mode=truncate_mode,
color_threshold=color_threshold,
get_leaves=get_leaves,
orientation=orientation,
labels=labels,
count_sort=count_sort,
distance_sort=distance_sort,
show_leaf_counts=show_leaf_counts,
i=ub, iv=iv + uwa, ivl=ivl, n=n,
icoord_list=icoord_list,
dcoord_list=dcoord_list, lvs=lvs,
current_color=current_color,
color_list=color_list,
currently_below_threshold=currently_below_threshold,
leaf_label_func=leaf_label_func,
level=level + 1, contraction_marks=contraction_marks,
link_color_func=link_color_func,
above_threshold_color=above_threshold_color,
traversal=traversal)
max_dist = max(uamd, ubmd, h)
icoord_list.append([uiva, uiva, uivb, uivb])
dcoord_list.append([uah, h, h, ubh])
if link_color_func is not None:
v = link_color_func(int(i))
if not isinstance(v, str):
raise TypeError("link_color_func must return a matplotlib "
"color string!")
color_list.append(v)
else:
color_list.append(c)
return (((uiva + uivb) / 2), uwa + uwb, h, max_dist)
def is_isomorphic(T1, T2):
"""
Determine if two different cluster assignments are equivalent.
Parameters
----------
T1 : array_like
An assignment of singleton cluster ids to flat cluster ids.
T2 : array_like
An assignment of singleton cluster ids to flat cluster ids.
Returns
-------
b : bool
Whether the flat cluster assignments `T1` and `T2` are
equivalent.
See Also
--------
linkage: for a description of what a linkage matrix is.
fcluster: for the creation of flat cluster assignments.
Examples
--------
>>> from scipy.cluster.hierarchy import fcluster, is_isomorphic
>>> from scipy.cluster.hierarchy import single, complete
>>> from scipy.spatial.distance import pdist
Two flat cluster assignments can be isomorphic if they represent the same
cluster assignment, with different labels.
For example, we can use the `scipy.cluster.hierarchy.single`: method
and flatten the output to four clusters:
>>> X = [[0, 0], [0, 1], [1, 0],
... [0, 4], [0, 3], [1, 4],
... [4, 0], [3, 0], [4, 1],
... [4, 4], [3, 4], [4, 3]]
>>> Z = single(pdist(X))
>>> T = fcluster(Z, 1, criterion='distance')
>>> T
array([3, 3, 3, 4, 4, 4, 2, 2, 2, 1, 1, 1], dtype=int32)
We can then do the same using the
`scipy.cluster.hierarchy.complete`: method:
>>> Z = complete(pdist(X))
>>> T_ = fcluster(Z, 1.5, criterion='distance')
>>> T_
array([1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4], dtype=int32)
As we can see, in both cases we obtain four clusters and all the data
points are distributed in the same way - the only thing that changes
are the flat cluster labels (3 => 1, 4 =>2, 2 =>3 and 4 =>1), so both
cluster assignments are isomorphic:
>>> is_isomorphic(T, T_)
True
"""
T1 = np.asarray(T1, order='c')
T2 = np.asarray(T2, order='c')
if type(T1) != np.ndarray:
raise TypeError('T1 must be a numpy array.')
if type(T2) != np.ndarray:
raise TypeError('T2 must be a numpy array.')
T1S = T1.shape
T2S = T2.shape
if len(T1S) != 1:
raise ValueError('T1 must be one-dimensional.')
if len(T2S) != 1:
raise ValueError('T2 must be one-dimensional.')
if T1S[0] != T2S[0]:
raise ValueError('T1 and T2 must have the same number of elements.')
n = T1S[0]
d1 = {}
d2 = {}
for i in range(0, n):
if T1[i] in d1:
if not T2[i] in d2:
return False
if d1[T1[i]] != T2[i] or d2[T2[i]] != T1[i]:
return False
elif T2[i] in d2:
return False
else:
d1[T1[i]] = T2[i]
d2[T2[i]] = T1[i]
return True
def maxdists(Z):
"""
Return the maximum distance between any non-singleton cluster.
Parameters
----------
Z : ndarray
The hierarchical clustering encoded as a matrix. See
``linkage`` for more information.
Returns
-------
maxdists : ndarray
A ``(n-1)`` sized numpy array of doubles; ``MD[i]`` represents
the maximum distance between any cluster (including
singletons) below and including the node with index i. More
specifically, ``MD[i] = Z[Q(i)-n, 2].max()`` where ``Q(i)`` is the
set of all node indices below and including node i.
See Also
--------
linkage: for a description of what a linkage matrix is.
is_monotonic: for testing for monotonicity of a linkage matrix.
Examples
--------
>>> from scipy.cluster.hierarchy import median, maxdists
>>> from scipy.spatial.distance import pdist
Given a linkage matrix ``Z``, `scipy.cluster.hierarchy.maxdists`
computes for each new cluster generated (i.e., for each row of the linkage
matrix) what is the maximum distance between any two child clusters.
Due to the nature of hierarchical clustering, in many cases this is going
to be just the distance between the two child clusters that were merged
to form the current one - that is, Z[:,2].
However, for non-monotonic cluster assignments such as
`scipy.cluster.hierarchy.median` clustering this is not always the
case: There may be cluster formations were the distance between the two
clusters merged is smaller than the distance between their children.
We can see this in an example:
>>> X = [[0, 0], [0, 1], [1, 0],
... [0, 4], [0, 3], [1, 4],
... [4, 0], [3, 0], [4, 1],
... [4, 4], [3, 4], [4, 3]]
>>> Z = median(pdist(X))
>>> Z
array([[ 0. , 1. , 1. , 2. ],
[ 3. , 4. , 1. , 2. ],
[ 9. , 10. , 1. , 2. ],
[ 6. , 7. , 1. , 2. ],
[ 2. , 12. , 1.11803399, 3. ],
[ 5. , 13. , 1.11803399, 3. ],
[ 8. , 15. , 1.11803399, 3. ],
[11. , 14. , 1.11803399, 3. ],
[18. , 19. , 3. , 6. ],
[16. , 17. , 3.5 , 6. ],
[20. , 21. , 3.25 , 12. ]])
>>> maxdists(Z)
array([1. , 1. , 1. , 1. , 1.11803399,
1.11803399, 1.11803399, 1.11803399, 3. , 3.5 ,
3.5 ])
Note that while the distance between the two clusters merged when creating the
last cluster is 3.25, there are two children (clusters 16 and 17) whose distance
is larger (3.5). Thus, `scipy.cluster.hierarchy.maxdists` returns 3.5 in
this case.
"""
Z = np.asarray(Z, order='c', dtype=np.double)
is_valid_linkage(Z, throw=True, name='Z')
n = Z.shape[0] + 1
MD = np.zeros((n - 1,))
[Z] = _copy_arrays_if_base_present([Z])
_hierarchy.get_max_dist_for_each_cluster(Z, MD, int(n))
return MD
def maxinconsts(Z, R):
"""
Return the maximum inconsistency coefficient for each
non-singleton cluster and its children.
Parameters
----------
Z : ndarray
The hierarchical clustering encoded as a matrix. See
`linkage` for more information.
R : ndarray
The inconsistency matrix.
Returns
-------
MI : ndarray
A monotonic ``(n-1)``-sized numpy array of doubles.
See Also
--------
linkage: for a description of what a linkage matrix is.
inconsistent: for the creation of a inconsistency matrix.
Examples
--------
>>> from scipy.cluster.hierarchy import median, inconsistent, maxinconsts
>>> from scipy.spatial.distance import pdist
Given a data set ``X``, we can apply a clustering method to obtain a
linkage matrix ``Z``. `scipy.cluster.hierarchy.inconsistent` can
be also used to obtain the inconsistency matrix ``R`` associated to
this clustering process:
>>> X = [[0, 0], [0, 1], [1, 0],
... [0, 4], [0, 3], [1, 4],
... [4, 0], [3, 0], [4, 1],
... [4, 4], [3, 4], [4, 3]]
>>> Z = median(pdist(X))
>>> R = inconsistent(Z)
>>> Z
array([[ 0. , 1. , 1. , 2. ],
[ 3. , 4. , 1. , 2. ],
[ 9. , 10. , 1. , 2. ],
[ 6. , 7. , 1. , 2. ],
[ 2. , 12. , 1.11803399, 3. ],
[ 5. , 13. , 1.11803399, 3. ],
[ 8. , 15. , 1.11803399, 3. ],
[11. , 14. , 1.11803399, 3. ],
[18. , 19. , 3. , 6. ],
[16. , 17. , 3.5 , 6. ],
[20. , 21. , 3.25 , 12. ]])
>>> R
array([[1. , 0. , 1. , 0. ],
[1. , 0. , 1. , 0. ],
[1. , 0. , 1. , 0. ],
[1. , 0. , 1. , 0. ],
[1.05901699, 0.08346263, 2. , 0.70710678],
[1.05901699, 0.08346263, 2. , 0.70710678],
[1.05901699, 0.08346263, 2. , 0.70710678],
[1.05901699, 0.08346263, 2. , 0.70710678],
[1.74535599, 1.08655358, 3. , 1.15470054],
[1.91202266, 1.37522872, 3. , 1.15470054],
[3.25 , 0.25 , 3. , 0. ]])
Here, `scipy.cluster.hierarchy.maxinconsts` can be used to compute
the maximum value of the inconsistency statistic (the last column of
``R``) for each non-singleton cluster and its children:
>>> maxinconsts(Z, R)
array([0. , 0. , 0. , 0. , 0.70710678,
0.70710678, 0.70710678, 0.70710678, 1.15470054, 1.15470054,
1.15470054])
"""
Z = np.asarray(Z, order='c')
R = np.asarray(R, order='c')
is_valid_linkage(Z, throw=True, name='Z')
is_valid_im(R, throw=True, name='R')
n = Z.shape[0] + 1
if Z.shape[0] != R.shape[0]:
raise ValueError("The inconsistency matrix and linkage matrix each "
"have a different number of rows.")
MI = np.zeros((n - 1,))
[Z, R] = _copy_arrays_if_base_present([Z, R])
_hierarchy.get_max_Rfield_for_each_cluster(Z, R, MI, int(n), 3)
return MI
def maxRstat(Z, R, i):
"""
Return the maximum statistic for each non-singleton cluster and its
children.
Parameters
----------
Z : array_like
The hierarchical clustering encoded as a matrix. See `linkage` for more
information.
R : array_like
The inconsistency matrix.
i : int
The column of `R` to use as the statistic.
Returns
-------
MR : ndarray
Calculates the maximum statistic for the i'th column of the
inconsistency matrix `R` for each non-singleton cluster
node. ``MR[j]`` is the maximum over ``R[Q(j)-n, i]``, where
``Q(j)`` the set of all node ids corresponding to nodes below
and including ``j``.
See Also
--------
linkage: for a description of what a linkage matrix is.
inconsistent: for the creation of a inconsistency matrix.
Examples
--------
>>> from scipy.cluster.hierarchy import median, inconsistent, maxRstat
>>> from scipy.spatial.distance import pdist
Given a data set ``X``, we can apply a clustering method to obtain a
linkage matrix ``Z``. `scipy.cluster.hierarchy.inconsistent` can
be also used to obtain the inconsistency matrix ``R`` associated to
this clustering process:
>>> X = [[0, 0], [0, 1], [1, 0],
... [0, 4], [0, 3], [1, 4],
... [4, 0], [3, 0], [4, 1],
... [4, 4], [3, 4], [4, 3]]
>>> Z = median(pdist(X))
>>> R = inconsistent(Z)
>>> R
array([[1. , 0. , 1. , 0. ],
[1. , 0. , 1. , 0. ],
[1. , 0. , 1. , 0. ],
[1. , 0. , 1. , 0. ],
[1.05901699, 0.08346263, 2. , 0.70710678],
[1.05901699, 0.08346263, 2. , 0.70710678],
[1.05901699, 0.08346263, 2. , 0.70710678],
[1.05901699, 0.08346263, 2. , 0.70710678],
[1.74535599, 1.08655358, 3. , 1.15470054],
[1.91202266, 1.37522872, 3. , 1.15470054],
[3.25 , 0.25 , 3. , 0. ]])
`scipy.cluster.hierarchy.maxRstat` can be used to compute
the maximum value of each column of ``R``, for each non-singleton
cluster and its children:
>>> maxRstat(Z, R, 0)
array([1. , 1. , 1. , 1. , 1.05901699,
1.05901699, 1.05901699, 1.05901699, 1.74535599, 1.91202266,
3.25 ])
>>> maxRstat(Z, R, 1)
array([0. , 0. , 0. , 0. , 0.08346263,
0.08346263, 0.08346263, 0.08346263, 1.08655358, 1.37522872,
1.37522872])
>>> maxRstat(Z, R, 3)
array([0. , 0. , 0. , 0. , 0.70710678,
0.70710678, 0.70710678, 0.70710678, 1.15470054, 1.15470054,
1.15470054])
"""
Z = np.asarray(Z, order='c')
R = np.asarray(R, order='c')
is_valid_linkage(Z, throw=True, name='Z')
is_valid_im(R, throw=True, name='R')
if type(i) is not int:
raise TypeError('The third argument must be an integer.')
if i < 0 or i > 3:
raise ValueError('i must be an integer between 0 and 3 inclusive.')
if Z.shape[0] != R.shape[0]:
raise ValueError("The inconsistency matrix and linkage matrix each "
"have a different number of rows.")
n = Z.shape[0] + 1
MR = np.zeros((n - 1,))
[Z, R] = _copy_arrays_if_base_present([Z, R])
_hierarchy.get_max_Rfield_for_each_cluster(Z, R, MR, int(n), i)
return MR
def leaders(Z, T):
"""
Return the root nodes in a hierarchical clustering.
Returns the root nodes in a hierarchical clustering corresponding
to a cut defined by a flat cluster assignment vector ``T``. See
the ``fcluster`` function for more information on the format of ``T``.
For each flat cluster :math:`j` of the :math:`k` flat clusters
represented in the n-sized flat cluster assignment vector ``T``,
this function finds the lowest cluster node :math:`i` in the linkage
tree Z, such that:
* leaf descendants belong only to flat cluster j
(i.e., ``T[p]==j`` for all :math:`p` in :math:`S(i)`, where
:math:`S(i)` is the set of leaf ids of descendant leaf nodes
with cluster node :math:`i`)
* there does not exist a leaf that is not a descendant with
:math:`i` that also belongs to cluster :math:`j`
(i.e., ``T[q]!=j`` for all :math:`q` not in :math:`S(i)`). If
this condition is violated, ``T`` is not a valid cluster
assignment vector, and an exception will be thrown.
Parameters
----------
Z : ndarray
The hierarchical clustering encoded as a matrix. See
`linkage` for more information.
T : ndarray
The flat cluster assignment vector.
Returns
-------
L : ndarray
The leader linkage node id's stored as a k-element 1-D array,
where ``k`` is the number of flat clusters found in ``T``.
``L[j]=i`` is the linkage cluster node id that is the
leader of flat cluster with id M[j]. If ``i < n``, ``i``
corresponds to an original observation, otherwise it
corresponds to a non-singleton cluster.
M : ndarray
The leader linkage node id's stored as a k-element 1-D array, where
``k`` is the number of flat clusters found in ``T``. This allows the
set of flat cluster ids to be any arbitrary set of ``k`` integers.
For example: if ``L[3]=2`` and ``M[3]=8``, the flat cluster with
id 8's leader is linkage node 2.
See Also
--------
fcluster: for the creation of flat cluster assignments.
Examples
--------
>>> from scipy.cluster.hierarchy import ward, fcluster, leaders
>>> from scipy.spatial.distance import pdist
Given a linkage matrix ``Z`` - obtained after apply a clustering method
to a dataset ``X`` - and a flat cluster assignment array ``T``:
>>> X = [[0, 0], [0, 1], [1, 0],
... [0, 4], [0, 3], [1, 4],
... [4, 0], [3, 0], [4, 1],
... [4, 4], [3, 4], [4, 3]]
>>> Z = ward(pdist(X))
>>> Z
array([[ 0. , 1. , 1. , 2. ],
[ 3. , 4. , 1. , 2. ],
[ 6. , 7. , 1. , 2. ],
[ 9. , 10. , 1. , 2. ],
[ 2. , 12. , 1.29099445, 3. ],
[ 5. , 13. , 1.29099445, 3. ],
[ 8. , 14. , 1.29099445, 3. ],
[11. , 15. , 1.29099445, 3. ],
[16. , 17. , 5.77350269, 6. ],
[18. , 19. , 5.77350269, 6. ],
[20. , 21. , 8.16496581, 12. ]])
>>> T = fcluster(Z, 3, criterion='distance')
>>> T
array([1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4], dtype=int32)
`scipy.cluster.hierarchy.leaders` returns the indices of the nodes
in the dendrogram that are the leaders of each flat cluster:
>>> L, M = leaders(Z, T)
>>> L
array([16, 17, 18, 19], dtype=int32)
(remember that indices 0-11 point to the 12 data points in ``X``,
whereas indices 12-22 point to the 11 rows of ``Z``)
`scipy.cluster.hierarchy.leaders` also returns the indices of
the flat clusters in ``T``:
>>> M
array([1, 2, 3, 4], dtype=int32)
"""
Z = np.asarray(Z, order='c')
T = np.asarray(T, order='c')
if type(T) != np.ndarray or T.dtype != 'i':
raise TypeError('T must be a one-dimensional numpy array of integers.')
is_valid_linkage(Z, throw=True, name='Z')
if len(T) != Z.shape[0] + 1:
raise ValueError('Mismatch: len(T)!=Z.shape[0] + 1.')
Cl = np.unique(T)
kk = len(Cl)
L = np.zeros((kk,), dtype='i')
M = np.zeros((kk,), dtype='i')
n = Z.shape[0] + 1
[Z, T] = _copy_arrays_if_base_present([Z, T])
s = _hierarchy.leaders(Z, T, L, M, int(kk), int(n))
if s >= 0:
raise ValueError(('T is not a valid assignment vector. Error found '
'when examining linkage node %d (< 2n-1).') % s)
return (L, M)Implementing Dendrogram Cuts in Plotly
Modification of create_dendrogram function and _Dendrogram class
To implement cuts, the basic modification to be made is including truncation parameters such as truncate_mode and p to all wrappers around the scipy’s dendrogram function. The following changes are made:
create_dendrogramnow includespandtruncate_modeas parametersthe plotly class
_Dendrogramnow has bothpandtruncate_modeas object parametersthe
get_dendrogram_tracesmethod now passes on these parameters to its call to scipy’sdendogramfunction
from __future__ import absolute_import
from collections import OrderedDict
from plotly import exceptions, optional_imports
from plotly.graph_objs import graph_objs
# Optional imports, may be None for users that only use our core functionality.
np = optional_imports.get_module("numpy")
scp = optional_imports.get_module("scipy")
sch = optional_imports.get_module("scipy.cluster.hierarchy")
scs = optional_imports.get_module("scipy.spatial")
def create_dendrogram(
X,
p=30,
truncate_mode=None,
orientation="bottom",
labels=None,
colorscale=None,
distfun=None,
linkagefun=lambda x: sch.linkage(x, "complete"),
hovertext=None,
color_threshold=None,
leaf_label_func=None,
):
"""
Function that returns a dendrogram Plotly figure object. This is a thin, modified
wrapper around scipy.cluster.hierarchy.dendrogram that includes truncation parameters.
See also https://dash.plot.ly/dash-bio/clustergram.
:param (ndarray) X: Matrix of observations as array of arrays
:param (str) orientation: 'top', 'right', 'bottom', or 'left'
:param (list) labels: List of axis category labels(observation labels)
:param (list) colorscale: Optional colorscale for the dendrogram tree.
Requires 8 colors to be specified, the 7th of
which is ignored. With scipy>=1.5.0, the 2nd, 3rd
and 6th are used twice as often as the others.
Given a shorter list, the missing values are
replaced with defaults and with a longer list the
extra values are ignored.
:param (function) distfun: Function to compute the pairwise distance from
the observations
:param (function) linkagefun: Function to compute the linkage matrix from
the pairwise distances
:param (list[list]) hovertext: List of hovertext for constituent traces of dendrogram
clusters
:param (double) color_threshold: Value at which the separation of clusters will be made
Example 1: Simple bottom oriented dendrogram
>>> from plotly.figure_factory import create_dendrogram
>>> import numpy as np
>>> X = np.random.rand(10,10)
>>> fig = create_dendrogram(X)
>>> fig.show()
Example 2: Dendrogram to put on the left of the heatmap
>>> from plotly.figure_factory import create_dendrogram
>>> import numpy as np
>>> X = np.random.rand(5,5)
>>> names = ['Jack', 'Oxana', 'John', 'Chelsea', 'Mark']
>>> dendro = create_dendrogram(X, orientation='right', labels=names)
>>> dendro.update_layout({'width':700, 'height':500}) # doctest: +SKIP
>>> dendro.show()
Example 3: Dendrogram with Pandas
>>> from plotly.figure_factory import create_dendrogram
>>> import numpy as np
>>> import pandas as pd
>>> Index= ['A','B','C','D','E','F','G','H','I','J']
>>> df = pd.DataFrame(abs(np.random.randn(10, 10)), index=Index)
>>> fig = create_dendrogram(df, labels=Index)
>>> fig.show()
"""
if not scp or not scs or not sch:
raise ImportError(
"FigureFactory.create_dendrogram requires scipy, \
scipy.spatial and scipy.hierarchy"
)
s = X.shape
if len(s) != 2:
exceptions.PlotlyError("X should be 2-dimensional array.")
if distfun is None:
distfun = scs.distance.pdist
dendrogram = _Dendrogram(
X=X,
p=p,
truncate_mode=truncate_mode,
orientation=orientation,
labels=labels,
colorscale=colorscale,
distfun=distfun,
linkagefun=linkagefun,
hovertext=hovertext,
color_threshold=color_threshold,
leaf_label_func=leaf_label_func,
)
return graph_objs.Figure(data=dendrogram.data, layout=dendrogram.layout)
class _Dendrogram(object):
"""Refer to FigureFactory.create_dendrogram() for docstring."""
def __init__(
self,
X,
p=30,
truncate_mode=None,
orientation="bottom",
labels=None,
colorscale=None,
width=np.inf,
height=np.inf,
xaxis="xaxis",
yaxis="yaxis",
distfun=None,
linkagefun=lambda x: sch.linkage(x, "complete"),
hovertext=None,
color_threshold=None,
leaf_label_func=None,
):
self.p = p
self.truncate_mode=truncate_mode
self.orientation=orientation
self.labels = labels
self.xaxis = xaxis
self.yaxis = yaxis
self.data = []
self.leaves = []
self.sign = {self.xaxis: 1, self.yaxis: 1}
self.layout = {self.xaxis: {}, self.yaxis: {}}
self.leaf_label_func = leaf_label_func
if self.orientation in ["left", "bottom"]:
self.sign[self.xaxis] = 1
else:
self.sign[self.xaxis] = -1
if self.orientation in ["right", "bottom"]:
self.sign[self.yaxis] = 1
else:
self.sign[self.yaxis] = -1
if distfun is None:
distfun = scs.distance.pdist
(dd_traces, xvals, yvals, ordered_labels, leaves) = self.get_dendrogram_traces(
X, colorscale, distfun, linkagefun, hovertext, color_threshold,
)
self.labels = ordered_labels
self.leaves = leaves
yvals_flat = yvals.flatten()
xvals_flat = xvals.flatten()
self.zero_vals = []
for i in range(len(yvals_flat)):
if yvals_flat[i] == 0.0 and xvals_flat[i] not in self.zero_vals:
self.zero_vals.append(xvals_flat[i])
if len(self.zero_vals) > len(yvals) + 1:
# If the length of zero_vals is larger than the length of yvals,
# it means that there are wrong vals because of the identicial samples.
# Three and more identicial samples will make the yvals of spliting
# center into 0 and it will accidentally take it as leaves.
l_border = int(min(self.zero_vals))
r_border = int(max(self.zero_vals))
correct_leaves_pos = range(
l_border, r_border + 1, int((r_border - l_border) / len(yvals))
)
# Regenerating the leaves pos from the self.zero_vals with equally intervals.
self.zero_vals = [v for v in correct_leaves_pos]
self.zero_vals.sort()
self.layout = self.set_figure_layout(width, height)
self.data = dd_traces
def get_color_dict(self, colorscale):
"""
Returns colorscale used for dendrogram tree clusters.
:param (list) colorscale: Colors to use for the plot in rgb format.
:rtype (dict): A dict of default colors mapped to the user colorscale.
"""
# These are the color codes returned for dendrograms
# We're replacing them with nicer colors
# This list is the colors that can be used by dendrogram, which were
# determined as the combination of the default above_threshold_color and
# the default color palette (see scipy/cluster/hierarchy.py)
d = {
"r": "red",
"g": "green",
"b": "blue",
"c": "cyan",
"m": "magenta",
"y": "yellow",
"k": "black",
# TODO: 'w' doesn't seem to be in the default color
# palette in scipy/cluster/hierarchy.py
"w": "white",
}
default_colors = OrderedDict(sorted(d.items(), key=lambda t: t[0]))
if colorscale is None:
rgb_colorscale = [
"rgb(0,116,217)", # blue
"rgb(35,205,205)", # cyan
"rgb(61,153,112)", # green
"rgb(40,35,35)", # black
"rgb(133,20,75)", # magenta
"rgb(255,65,54)", # red
"rgb(255,255,255)", # white
"rgb(255,220,0)", # yellow
]
else:
rgb_colorscale = colorscale
for i in range(len(default_colors.keys())):
k = list(default_colors.keys())[i] # PY3 won't index keys
if i < len(rgb_colorscale):
default_colors[k] = rgb_colorscale[i]
# add support for cyclic format colors as introduced in scipy===1.5.0
# before this, the colors were named 'r', 'b', 'y' etc., now they are
# named 'C0', 'C1', etc. To keep the colors consistent regardless of the
# scipy version, we try as much as possible to map the new colors to the
# old colors
# this mapping was found by inpecting scipy/cluster/hierarchy.py (see
# comment above).
new_old_color_map = [
("C0", "b"),
("C1", "g"),
("C2", "r"),
("C3", "c"),
("C4", "m"),
("C5", "y"),
("C6", "k"),
("C7", "g"),
("C8", "r"),
("C9", "c"),
]
for nc, oc in new_old_color_map:
try:
default_colors[nc] = default_colors[oc]
except KeyError:
# it could happen that the old color isn't found (if a custom
# colorscale was specified), in this case we set it to an
# arbitrary default.
default_colors[n] = "rgb(0,116,217)"
return default_colors
def set_axis_layout(self, axis_key):
"""
Sets and returns default axis object for dendrogram figure.
:param (str) axis_key: E.g., 'xaxis', 'xaxis1', 'yaxis', yaxis1', etc.
:rtype (dict): An axis_key dictionary with set parameters.
"""
axis_defaults = {
"type": "linear",
"ticks": "outside",
"mirror": "allticks",
"rangemode": "tozero",
"showticklabels": True,
"zeroline": False,
"showgrid": False,
"showline": True,
}
if len(self.labels) != 0:
axis_key_labels = self.xaxis
if self.orientation in ["left", "right"]:
axis_key_labels = self.yaxis
if axis_key_labels not in self.layout:
self.layout[axis_key_labels] = {}
self.layout[axis_key_labels]["tickvals"] = [
zv * self.sign[axis_key] for zv in self.zero_vals
]
self.layout[axis_key_labels]["ticktext"] = self.labels
self.layout[axis_key_labels]["tickmode"] = "array"
self.layout[axis_key].update(axis_defaults)
return self.layout[axis_key]
def set_figure_layout(self, width, height):
"""
Sets and returns default layout object for dendrogram figure.
"""
self.layout.update(
{
"showlegend": False,
"autosize": False,
"hovermode": "closest",
"width": width,
"height": height,
}
)
self.set_axis_layout(self.xaxis)
self.set_axis_layout(self.yaxis)
return self.layout
def get_dendrogram_traces(
self, X, colorscale, distfun, linkagefun, hovertext, color_threshold
):
"""
Calculates all the elements needed for plotting a dendrogram.
:param (ndarray) X: Matrix of observations as array of arrays
:param (list) colorscale: Color scale for dendrogram tree clusters
:param (function) distfun: Function to compute the pairwise distance
from the observations
:param (function) linkagefun: Function to compute the linkage matrix
from the pairwise distances
:param (list) hovertext: List of hovertext for constituent traces of dendrogram
:rtype (tuple): Contains all the traces in the following order:
(a) trace_list: List of Plotly trace objects for dendrogram tree
(b) icoord: All X points of the dendrogram tree as array of arrays
with length 4
(c) dcoord: All Y points of the dendrogram tree as array of arrays
with length 4
(d) ordered_labels: leaf labels in the order they are going to
appear on the plot
(e) P['leaves']: left-to-right traversal of the leaves
"""
d = distfun(X)
Z = linkagefun(d)
P = sch.dendrogram(
Z,
p=self.p,
truncate_mode=self.truncate_mode,
orientation=self.orientation,
labels=self.labels,
no_plot=True,
color_threshold=color_threshold,
leaf_label_func=self.leaf_label_func
)
icoord = np.array(P["icoord"])
dcoord = np.array(P["dcoord"])
ordered_labels = np.array(P["ivl"])
color_list = np.array(P["color_list"])
colors = self.get_color_dict(colorscale)
trace_list = []
for i in range(len(icoord)):
# xs and ys are arrays of 4 points that make up the '∩' shapes
# of the dendrogram tree
if self.orientation in ["top", "bottom"]:
xs = icoord[i]
else:
xs = dcoord[i]
if self.orientation in ["top", "bottom"]:
ys = dcoord[i]
else:
ys = icoord[i]
color_key = color_list[i]
hovertext_label = None
if hovertext:
hovertext_label = hovertext[i]
trace = dict(
type="scatter",
x=np.multiply(self.sign[self.xaxis], xs),
y=np.multiply(self.sign[self.yaxis], ys),
mode="lines",
marker=dict(color=colors[color_key]),
text=hovertext_label,
hoverinfo="text",
)
try:
x_index = int(self.xaxis[-1])
except ValueError:
x_index = ""
try:
y_index = int(self.yaxis[-1])
except ValueError:
y_index = ""
trace["xaxis"] = "x" + x_index
trace["yaxis"] = "y" + y_index
trace_list.append(trace)
return trace_list, icoord, dcoord, ordered_labels, P["leaves"]Creating Hovertext Robust to Truncation
In Dendrograms with hovertext ranging a few hundred nodes, truncation is necessary
The Z matrix contains contains \(n-1\) rows each representing new clusters, where \(n\) is then number of data points.
- higher rows signify larger clusters, ordered by the distance metric of their child clusters.
- the row indices also correspond to the
ClusterNodeinstances in the node list returned byscipy.cluster.hierarchy.to_tree - default indexing will have the root cluster as \(n-2\)
Truncation when
truncate_modeis set to ‘lastp’ is governed by parameterppbecomes the number of leaves to be displayed- there will be a total of \(p - 1\) new clusters displayed
- the node indices will then be in range [2n-2-p, 2n-2], or
range(2*n - 2 - p, 2*n - 1)
If indices are shown
For UI considerations, a shift in
pSimpler indices, \(j\) can be transformations of default indices \(i\) for UI considerations can be given by: \[j = 2n-2 - i\]
Visualization Techniques
t-Stochastic Neighbor Embedding (TSNE)
This is a technique that maps a higher dimensional space to a lower dimensional space while preserving probabilities of choosing analogous neighboring points between the two difference spaces.
The original paper can be read here as well as a post about effective usage.
Uniform Manifold Approximation (UMAP)
This is another method of dimensional reduction but works well for non-linear mappings.
Implementation with Plotly
Here is an example of using Plotly to visualize data sets using both techniques, straight from their documentation
import plotly.express as px
df = px.data.iris()
features = ["sepal_width", "sepal_length", "petal_width", "petal_length"]
fig = px.scatter_matrix(df, dimensions=features, color="species")
fig.show()
from sklearn.manifold import TSNE, trustworthiness
import plotly.express as px
df = px.data.iris()
features = df.loc[:, :'petal_width']
tsne = TSNE(n_components=2, random_state=0)
projections = tsne.fit_transform(features)
fig = px.scatter(
projections, x=0, y=1,
color=df.species, labels={'color': 'species'}
)
fig.show()
score = trustworthiness(features.to_numpy(), projections, n_neighbors=5)
print(f'Trustworthiness score is: {score}')Trustworthiness score is: 0.9880657276995305
from umap import UMAP
import plotly.express as px
df = px.data.iris()
features = df.loc[:, :'petal_width']
umap_2d = UMAP(random_state=0, n_components=2)
umap_3d = UMAP(random_state=0, n_components=3, n_neighbors=5)
proj_2d = umap_2d.fit_transform(features)
fig_2d = px.scatter(proj_2d, x=0, y=1,
color=df.species, labels={'color': 'species'}
)
fig_2d.show()C:\Users\Jonathan\anaconda3\envs\py313\Lib\site-packages\umap\umap_.py:1952: UserWarning:
n_jobs value 1 overridden to 1 by setting random_state. Use no seed for parallelism.

score = trustworthiness(features.to_numpy(), projections, n_neighbors=5)
print(f'Trustworthiness score is: {score}')Trustworthiness score is: 0.9880657276995305